{"id":35058,"date":"2022-08-25T10:06:50","date_gmt":"2022-08-25T04:36:50","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35058"},"modified":"2022-08-25T10:06:50","modified_gmt":"2022-08-25T04:36:50","slug":"inter-1st-year-maths-1a-trigonometric-equations-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1a-trigonometric-equations-important-questions\/","title":{"rendered":"Inter 1st Year Maths 1A Trigonometric Equations Important Questions"},"content":{"rendered":"

Students get through Maths 1A Important Questions<\/a> Inter 1st Year Maths 1A Trigonometric Equations Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 1st Year Maths 1A Trigonometric Equations Important Questions<\/h2>\n

Question 1.
\nSolve sin x = \\(\\frac{1}{\\sqrt{2}}\\)
\nSolution:
\nsin x = \\(\\frac{1}{\\sqrt{2}}\\) = sin \\(\\frac{\\pi}{4}\\) and \\(\\frac{\\pi}{4}\\) \u2208 [-\\(\\frac{\\pi}{2}\\), \\(\\frac{\\pi}{2}\\)]
\n\u2234 x = \\(\\frac{\\pi}{4}\\) is the principal solution and
\nx = n\u03c0 + (-1)n<\/sup> \\(\\frac{\\pi}{4}\\), n \u2208 Z is the general solution.<\/p>\n

\"Inter<\/p>\n

Question 2.
\nSolve sin 2\u03b8 = \\(\\frac{\\sqrt{5}-1}{4}\\)
\nSolution:
\nsin 2\u03b8 = \\(\\frac{\\sqrt{5}-1}{4}\\) = sin 18\u00b0 = sin(\\(\\frac{\\pi}{10}\\)) and \\(\\frac{\\pi}{10}\\) \u2208 [-\\(\\frac{\\pi}{2}\\), \\(\\frac{\\pi}{2}\\)]
\n\u21d2 2\u03b8 = \\(\\frac{\\pi}{10}\\) and hence \u03b8 = \\(\\frac{\\pi}{20}\\) is the principal solution.
\nGeneral solution is given by
\n2\u03b8 = n\u03c0 + (-1)n<\/sup> \\(\\frac{\\pi}{10}\\), n \u2208 Z (or)
\n\u03b8 = n\\(\\frac{\\pi}{2}\\) + (-1)n<\/sup> \\(\\frac{\\pi}{20}\\), n \u2208 Z<\/p>\n

Question 3.
\nSolve tan2<\/sup>\u03b8 = 3
\nSolution:
\ntan2<\/sup>\u03b8 = 3 = tan \u03b8 = \u00b1\\(\\sqrt{3}\\) = tan (\u00b1\\(\\frac{\\pi}{3}\\))
\nand \u00b1 \\(\\frac{\\pi}{3}\\) \u2208 (-\\(\\frac{\\pi}{2}\\), \\(\\frac{\\pi}{2}\\))
\n\u2234 \u03b8 = \u00b1 \\(\\frac{\\pi}{3}\\) are the principal solutions of the given equation.
\nGeneral solution is given by n\u03c0 \u00b1 \\(\\frac{\\pi}{3}\\), n \u2208 Z<\/p>\n

Question 4.
\nSolve 3 cosec x = 4 sin x
\nSolution:
\n3 cosec x = 4 sin x
\n\u21d4 \\(\\frac{3}{\\sin x}\\) = 4 sin x
\n\u21d2 sin2<\/sup>x = \\(\\frac{3}{4}\\)
\n\u21d4 sin x = \u00b1 \\(\\frac{\\sqrt{3}}{2}\\)
\n\u2234 Principal solutions are x = \u00b1 \\(\\frac{\\pi}{3}\\)
\nGeneral solutions are given by x = n\u03c0 \u00b1 \\(\\frac{\\pi}{3}\\), n \u2208 Z<\/p>\n

Question 5.
\nIf x is acute and sin (x + 10\u00b0) = cos (3x – 68\u00b0), find x in degrees.
\nSolution:
\nGiven sin (x + 10\u00b0) = cos (3x – 68\u00b0)
\n\u21d4 sin(x + 10\u00b0) = sin (90\u00b0 + 3x – 68\u00b0) = sin (22\u00b0 + 3x)
\n\u2234 x + 10\u00b0 = n(180\u00b0) + (-1 )n<\/sup> (22\u00b0 + 3x)
\nIf n = 2k, k \u2208 Z then
\nx + 10\u00b0 = (2k) (180\u00b0) + (220 + 3x)
\n\u21d2 2x = -k(360\u00b0) – 12\u00b0
\n\u21d2 x = \\(\\frac{-k\\left(360^{\\circ}\\right)-12^{\\circ}}{2}\\)
\n= -k(180\u00b0) – 6\u00b0
\nThis is not valid because for any integral k, x is not acute
\nIf n = 2k + 1, then
\nx + 10\u00b0 = (2k + 1) 180\u00b0 – (22\u00b0+ 3x)
\n\u21d2 4x = (2k + 1) 180\u00b0 – 32\u00b0
\n\u21d2 x = (2k + 1) 45\u00b0 – 8\u00b0
\nNow k = 0 \u21d2 x = 37\u00b0
\nFor other integral values of k, x is not acute
\n\u2234 The only solution is x = 37\u00b0<\/p>\n

Question 6.
\nSolve cos 3\u03b8 = sin 2\u03b8
\nSolution:
\ncos 3\u03b8 = sin 2\u03b8 = cas (\\(\\frac{\\pi}{2}\\) – 2\u03b8)
\n\u21d2 3\u03b8 = 2n\u03c0 \u00b1 (\\(\\frac{\\pi}{2}\\) – 2\u03b8), n \u2208 Z
\n\u21d2 3\u03b8 = 2n\u03c0 + (\\(\\frac{\\pi}{2}\\) – 2\u03b8) (or)
\n3\u03b8 = 2n\u03c0 – (\\(\\frac{\\pi}{2}\\) – 2\u03b8)
\n\u21d2 5\u03b8 = 2n\u03c0 + \\(\\frac{\\pi}{2}\\) (or) \u03b8 = 2n\u03c0 – \\(\\frac{\\pi}{2}\\)
\n\u21d2 \u03b8 = (4n + 1) \\(\\frac{\\pi}{10}\\), n \u2208 Z (or)
\n\u03b8 = (4n – 1) \\(\\frac{\\pi}{2}\\), n \u2208 Z<\/p>\n

Question 7.
\nSolve 7 sin2<\/sup>\u03b8 + 3cos2<\/sup>\u03b8 = 4
\nSolution:
\nGiven that
\n7 sin2<\/sup>\u03b8 + 3 cos2<\/sup>\u03b8 = 4
\n\u21d2 7 sin2<\/sup>\u03b8 + 3(1 – sin2<\/sup>\u03b8) = 4
\n\u21d2 4sin2<\/sup>\u03b8 = 1
\n\u21d2 sin \u03b8 = \u00b1 \\(\\frac{1}{2}\\)
\n\u2234 Principal solutions are \u03b8 = \u00b1 \\(\\frac{\\pi}{6}\\) and the general solution is given by
\n\u03b8 = n\u03c0 \u00b1 \\(\\frac{\\pi}{6}\\), n \u2208 z.<\/p>\n

Question 8.
\nSolve 2 cos2<\/sup>\u03b8 – \\(\\sqrt{3}\\) sin \u03b8 + 1 = 0
\nSolution:
\n2 cos2<\/sup>\u03b8 – \\(\\sqrt{3}\\) sin \u03b8 + 1 = 0
\n\u21d2 2(1 – sin2<\/sup>\u03b8) – \\(\\sqrt{3}\\) sin \u03b8 + 1 = 0
\n\u21d2 2 sin2<\/sup>\u03b8 + \\(\\sqrt{3}\\) sin \u03b8 – 3 = 0
\n\u21d2 (2 sin \u03b8 – \\(\\sqrt{3}\\))(sin \u03b8 + \\(\\sqrt{3}\\)) = 0
\n\u21d2 sin \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\) and sin \u03b8 \u2260 \\(\\sqrt{3}\\)
\n\u2234 sin \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\) = sin (\\(\\frac{\\pi}{3}\\))
\n\u2234 \u03b8 = \\(\\frac{\\pi}{3}\\) is the principal solution and general solution is given by
\n\u03b8 = n\u03c0 + (-1)n<\/sup> \\(\\frac{\\pi}{3}\\), n \u2208 Z.<\/p>\n

Question 9.
\nFind all values of x \u2260 0 in (-\u03c0, \u03c0) satisfying the equation 81+cosx+cos2x+……<\/sup> = 43<\/sup>.
\nSolution:
\nIf cos x = \u00b11, the sum
\n1 + cos x + cos2<\/sup> x + …………… \u221e does not converse.
\nAssume |cos x| < 1.
\nThen 1 + cos x + cos2<\/sup> x + …………… \u221e
\n= \\(\\frac{1}{1-\\cos x}\\)
\n(\u2235 a + ar + ar2<\/sup> + ……………. \u221e = \\(\\frac{a}{1-r}\\), if |r| < 1)
\nNow 81+cosx+cos2x+………….<\/sup> = 43<\/sup>
\n\u21d2 (8)\\(\\frac{1}{1-\\cos x}\\)<\/sup> = 82<\/sup>
\n\u21d4 \\(\\frac{1}{1-\\cos x}\\) = 2 = cos x = \\(\\frac{1}{2}\\)
\n\u21d4 x = \\(\\frac{\\pi}{3}\\) or –\\(\\frac{\\pi}{3}\\) (\u2235 x \u2208 (-\u03c0, \u03c0))<\/p>\n

Question 10.
\nSolve tan \u03b8 + 3 cot \u03b8 = 5 sec \u03b8
\nSolution:
\ntan \u03b8 + 3 cot \u03b8 = 5 sec \u03b8 is valid only when
\ncos \u03b8 \u2260 0 and sin \u03b8 \u2260 0
\n\u21d2 \\(\\frac{\\sin \\theta}{\\cos \\theta}\\) + 3\\(\\frac{\\cos \\theta}{\\sin \\theta}\\) =\\(\\frac{5}{\\cos \\theta}\\)
\n\u21d2 sin2<\/sup>\u03b8+ 3cos2<\/sup>\u03b8= 5 sin \u03b8
\n\u21d2 sin2<\/sup>\u03b8 + 3(1 – sin2<\/sup>\u03b8) – 5 sin \u03b8 = 0
\n\u21d2 2 sin2<\/sup>\u03b8 + 5 sin \u03b8 – 3 = 0
\n\u21d2 2 sin2<\/sup>\u03b8 + 6sin \u03b8 – sin \u03b8 – 3 = 0
\n\u21d2 2 sin \u03b8(sin \u03b8 + 3) – 1 (sin \u03b8 + 3)= 0
\n\u21d2 (2 sin \u03b8 – 1)(sin \u03b8 + 3) = 0
\n\u21d2 sin \u03b8 = \\(\\frac{1}{2}\\) (\u2235sin \u03b8 \u2260 -3)
\n\u2234 sin = \\(\\frac{1}{2}\\) = sin (\\(\\frac{\\pi}{6}\\))
\n\u2234 Principal solution is \u03b8 = \\(\\frac{\\pi}{6}\\) and general solution is n\u03c0 + (-1)n<\/sup>\\(\\frac{\\pi}{6}\\), n \u2208 Z.<\/p>\n

Question 11.
\nSolve 1 + sin 2<\/sup>\u03b8 = 3 sin \u03b8 cos \u03b8. [Mar 11]
\nSolution:
\n1 + sin2<\/sup>\u03b8 = 3 sin \u03b8 cas \u03b8
\nDividing by cos2<\/sup>\u03b8, \u2235 cos \u03b8 \u2260 0
\nsec2<\/sup>\u03b8 + tan2<\/sup>\u03b8 = 3 tan \u03b8
\n\u21d2 (1 + tan2<\/sup>\u03b8) + tan2<\/sup>\u03b8 – 3 tan \u03b8 = 0
\n\u21d2 2tan2<\/sup>\u03b8 – 3 tan\u03b8 + 1 = 0
\n\u21d2 2tan2<\/sup>\u03b8 – 2 tan \u03b8 – tan \u03b8 + 1 = 0
\n\u21d2 2 tan \u03b8 (tan \u03b8 – 1) – (tan \u03b8 – 1) = 0
\n\u21d2 (tan \u03b8 – 1)(2 tan \u03b8 – 1) = 0
\n\u2234 tan \u03b8 = 1 (or) tan \u03b8 = \\(\\frac{1}{2}\\)
\nNow tan \u03b8 = 1 = tan \\(\\frac{\\pi}{4}\\)
\n\u2234 Principal solution \u03b8 = \\(\\frac{\\pi}{4}\\) and general solution is given by
\n\u03b8 = n\u03c0 + \\(\\frac{\\pi}{6}\\), n \u2208 Z
\nlet \u03b1 be the principal solution of tan \u03b8 = \\(\\frac{1}{2}\\)
\nThen the general solution is given by \u03b8 = n\u03c0 + \u03b1, n \u2208 Z<\/p>\n

Question 12.
\nSolve \\(\\sqrt{2}\\) (sin x + cos x) = \\(\\sqrt{3}\\) (A.P) [Mar. 16, 15, May 12, 08]
\nSolution:
\nGiven that \\(\\sqrt{2}\\) (sin x + cos x) = \\(\\sqrt{3}\\)
\n\u21d4 sin x + cos x = \\(\\sqrt{\\frac{3}{2}}\\)
\nBy dividing both sides by \\(\\sqrt{2}\\), we get
\n\\(\\frac{1}{\\sqrt{2}}\\) sin x + \\(\\frac{1}{\\sqrt{2}}\\) cos x = \\(\\sqrt{\\frac{3}{2}}\\)
\n\u21d2 sin x. sin \\(\\frac{\\pi}{4}\\) + cos x. cos \\(\\frac{\\pi}{4}\\) = \\(\\sqrt{\\frac{3}{2}}\\)
\n\u21d2 cos (x – \\(\\frac{\\pi}{4}\\)) = \\(\\sqrt{\\frac{3}{2}}\\) = cos (\\(\\frac{\\pi}{6}\\))
\nPrincipal solution is x – \\(\\frac{\\pi}{4}\\) = \\(\\frac{\\pi}{6}\\) (i.e.,) x = \\(\\frac{5 \\pi}{12}\\)
\nGeneral solution is given by
\nx – \\(\\frac{\\pi}{4}\\) = 2n\u03c0 \u00b1 \\(\\frac{\\pi}{6}\\), n \u2208 Z
\n\u21d2 x = 2n\u03c0 + latex]\\frac{5 \\pi}{12}[\/latex] (or) x = 2n\u03c0 + \\(\\frac{\\pi}{12}\\), n \u2208 Z<\/p>\n

Question 13.
\nFind general solution of \u03b8 which satisfies both the equations sin\u03b8 = –\\(\\frac{1}{2}\\) and cos\u03b8 = –\\(\\frac{\\sqrt{3}}{2}\\)
\nSolution:
\nGiven sin \u03b8 = –\\(\\frac{1}{2}\\) = – sin \\(\\frac{\\pi}{6}\\)
\n= sin (\u03c0 + \\(\\frac{\\pi}{6}\\)) = sin (2\u03c0 – \\(\\frac{\\pi}{6}\\))
\n= sin \\(\\frac{7\\pi}{6}\\) = sin \\(\\frac{11\\pi}{6}\\)
\n\u2234 Considering only angles in (0, 2\u03c0), the only values of \u03b8 satisfying sin \u03b8 = –\\(\\frac{1}{2}\\) are \\(\\frac{7\\pi}{6}\\) or \\(\\frac{11\\pi}{6}\\)
\ncos \u03b8 = –\\(\\frac{\\sqrt{3}}{2}\\) = – cos\\(\\frac{\\pi}{6}\\)
\n= cos (\u03c0 – \\(\\frac{\\pi}{6}\\)) or cos (\u03c0 + \\(\\frac{\\pi}{6}\\))
\n= cos \\(\\frac{5\\pi}{6}\\) or cos \\(\\frac{7\\pi}{6}\\).
\n\u2234 Considering only angles in (0, 2\u03c0), the only values of \u03b8 satisfying cos \u03b8 = –\\(\\frac{\\sqrt{3}}{2}\\) are \\(\\frac{5 \\pi}{6}\\) or \\(\\frac{7 \\pi}{6}\\).
\nThus \\(\\frac{7 \\pi}{6}\\) is the only angle which satisfies both the equations.
\nHence general solution for \u03b8 is
\n\u03b8 = 2n\u03c0 + \\(\\frac{7 \\pi}{6}\\), n \u2208 Z.<\/p>\n

Question 14.
\nIf \u03b81<\/sub>, \u03b82<\/sub> are solutions of the equation a cos 2\u03b8 + b sin 2\u03b8 = c, tan \u03b81<\/sub> \u2260 tan \u03b82<\/sub> and a + c \u2260 0. find the values of
\n(i) tan \u03b81<\/sub> + tan \u03b82<\/sub>
\n(ii) tan \u03b81<\/sub> . tan\u03b82<\/sub> (T.S.) [Mar 16, 15]
\nSolution:
\na cos 2\u03b8 + b sin 2\u03b8 = c
\n\u21d4 a(\\(\\frac{1-\\tan ^{2} \\theta}{1+\\tan ^{2} \\theta}\\)) + b(\\(\\frac{2 \\tan \\theta}{1+\\tan ^{2} \\theta}\\)) = c
\n\u21d4 a(1 – tan2<\/sup>\u03b8) + 2b tan \u03b8) = c(1 + tan2<\/sup>\u03b8)
\n(c + a) tan2<\/sup>\u03b8 – 2b tan \u03b8 + (c – a) = 0
\nThis is a quadratic equation in tan \u03b8.
\nIt has two roots tan \u03b81<\/sub> and tan \u03b82<\/sub>. since \u03b81<\/sub> and \u03b82<\/sub> are solutions for the given equation
\nSum of the roots = tan \u03b81<\/sub> + tan \u03b82<\/sub> = \\(\\frac{2 b}{a+c}\\)
\nProduct of the roots = tan \u03b81<\/sub> tan \u03b82<\/sub> = \\(\\frac{c-a}{c+a}\\)
\nNow tan(\u03b81<\/sub> + \u03b82<\/sub>) = \\(\\frac{\\tan \\theta_{1}+\\tan \\theta_{2}}{1-\\tan \\theta_{1} \\tan \\theta_{2}}\\)
\n= \\(\\frac{\\left(\\frac{2 b}{a+c}\\right)}{1-\\left(\\frac{c-a}{c+a}\\right)}\\) = \\(\\frac{2 \\mathrm{~b}}{2 \\mathrm{a}}\\) = \\(\\frac{b}{a}\\)<\/p>\n

Question 15.
\nSolve 4 sin x sin 2x sin 4x = sin 3x
\nSolution:
\nsin 3x = 4 sin x sin 2x sin 4x
\n= 2 sin x (2 sin 4x sin 2x)
\n= 2 sin x [cos (2x) – cos 6x]
\n\u21d4 sin 3x = 2 cos 2xsinx – 2 cos6x sin x
\n\u21d4 sin 3x = sin (3x) – sin x – 2 cos 6x sin x
\n\u21d2 2 cos 6x sin x + sin x = 0
\n\u21d2 sin x(2 cos 6x + 1) = 0
\n\u21d2 sin x = 0 (or) cos 6x = \\(\\frac{-1}{2}\\)
\nCase (i) : sin x = 0
\n\u21d2 x = 0 is the principal solution and x = n\u03c0, n \u2208 Z is the general solution.<\/p>\n

Case (ii) : cos 6x = \\(\\frac{-1}{2}\\) = cos (\\(\\frac{2 \\pi}{3}\\))
\n\u21d2 6x = \\(\\frac{2 \\pi}{3}\\) \u21d2 x = \\(\\frac{\\pi}{9}\\) is the principal solution
\nThe general solution is given by
\n6x = 2n\u03c0 \u00b1 \\(\\frac{2 \\pi}{3}\\), n \u2208 Z
\n\u21d2 x = \\(\\frac{n \\pi}{3}\\) \u00b1 \\(\\frac{\\pi}{9}\\), n \u2208 Z<\/p>\n

Question 16.
\nIf 0 < \u03b8 < \u03c0, solve cos \u03b8.cos 2\u03b8 cos 3\u03b8 = \\(\\frac{1}{4}\\).
\nSolution:
\n4 cos \u03b8 cos 2\u03b8 cos 3\u03b8 = 1
\n\u21d2 2 cos 2\u03b8 (2 cos 3\u03b8. cos \u03b8) = 1
\n\u21d2 2 cos 2\u03b8 (cos 4\u03b8 + cos 2\u03b8) = 1
\n\u21d2 2 cos 4\u03b8 cos 2\u03b8 + (2 cos2<\/sup> 2\u03b8 – 1) = 0
\n\u21d2 2 cos 4\u03b8 cos 2\u03b8 + cos 4\u03b8 = 0
\n\u21d2 cos 4\u03b8 (2 cos 2\u03b8 + 1) = 0
\n\u21d2 cos 4\u03b8 = 0 (or) cos 2\u03b8 = \\(\\frac{-1}{2}\\)
\nCase(i): cos 4\u03b8 = 0 = cos (\\(\\frac{\\pi}{2}\\))
\n\u21d2 4\u03b8 = \\(\\frac{\\pi}{2}\\) is the principal solution (or)
\n\u03b8 = \\(\\frac{\\pi}{8}\\).
\nThe general solution is given by
\n4\u03b8 = (2n + 1) \\(\\frac{\\pi}{2}\\), n \u2208 Z
\n\u21d2 \u03b8 = (2n + 1) \\(\\frac{\\pi}{8}\\), n \u2208 Z
\nPut n = 0, 1, 2, ……… we get
\n\\(\\left\\{\\frac{\\pi}{8}, \\frac{3 \\pi}{8}, \\frac{5 \\pi}{8}, \\frac{7 \\pi}{8}\\right\\}\\) are the solutions that lie in (0, \u03c0)
\nCase (ii) : cos 2\u03b8 = \\(\\frac{-1}{2}\\) = cos (\\(\\frac{2\\pi}{3}\\))
\n\u21d2 2\u03b8 = \\(\\frac{2\\pi}{3}\\) is the principal solution
\n(i.e.,) \u03b8 = \\(\\frac{\\pi}{3}\\)
\nThe general solution is given by
\n2\u03b8 = 2n\u03c0 \u00b1 \\(\\frac{2\\pi}{3}\\), n \u2208 Z
\n\u21d2 \u03b8 = n\u03c0 \u00b1 latex]\\frac{\\pi}{3}[\/latex], n \u2208 Z
\nPut n = 0, 1, we get \\(\\left\\{\\frac{\\pi}{3}, \\frac{2 \\pi}{3}\\right\\}\\) are the solutions that lie in the interval (0, \u03c0).
\nHence the solutions of the given equation in (0, \u03c0) are \\(\\left\\{\\frac{\\pi}{8}, \\frac{\\pi}{3}, \\frac{3 \\pi}{8}, \\frac{5 \\pi}{8}, \\frac{2 \\pi}{3}, \\frac{7 \\pi}{8}\\right\\}\\).<\/p>\n

Question 17.
\nGiven p \u2260 \u00b1 q, show that the solutions of cos p\u03b8 + cos q\u03b8 = 0 form two series each of which is in A.R Find also the common difference of each A.P.
\nSolution:
\ncos p\u03b8 + cos q\u03b8 = 0
\n\"Inter
\n\"Inter<\/p>\n

Question 18.
\nSolve sin 2x – cos 2x = sin x – cos x.
\nSolution:
\nsin 2x – cos 2x = sin x – cos x.
\n\u21d2 sin 2x – sin x = cos 2x – cos x
\n\"Inter
\nThe general solution is given by
\n\\(\\frac{3 x}{2}\\) = n\u03c0 + (\\(\\frac{-\\pi}{4}\\)), n \u2208 Z
\n\u21d2 x = \\(\\frac{2 n \\pi}{3}\\) – \\(\\frac{\\pi}{6}\\), n \u2208 Z
\n\u2234 Solution set for the given equation is {2n\u03c0 \/ n \u2208 Z} \u222a {\\(\\frac{2 n \\pi}{3}\\) – \\(\\frac{\\pi}{6}\\) \/ n \u2208 Z}<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Trigonometric Equations Important Questions which are most likely to be asked in the exam. Intermediate 1st Year Maths 1A Trigonometric Equations Important Questions Question 1. Solve sin x = Solution: sin x = = sin and \u2208 [-, ] \u2234 x = is …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35058"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35058"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35058\/revisions"}],"predecessor-version":[{"id":35059,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35058\/revisions\/35059"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35058"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35058"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35058"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}