{"id":10103,"date":"2024-02-27T09:23:10","date_gmt":"2024-02-27T03:53:10","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=10103"},"modified":"2024-02-28T17:46:14","modified_gmt":"2024-02-28T12:16:14","slug":"ts-inter-1st-year-maths-1a-trigonometric-equations-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1a-trigonometric-equations-important-questions\/","title":{"rendered":"TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions"},"content":{"rendered":"

Students must practice these\u00a0TS Inter 1st Year Maths 1A Important Questions<\/a> Chapter 7 Trigonometric Equations to help strengthen their preparations for exams.<\/p>\n

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

Question 1.
\nSolve 2cos2<\/sup>\u03b8 – \\(\\sqrt{3}\\) sin \u03b8+1 = 0
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\nSolve 1 +sin2<\/sup>\u03b8 = 3 sin\u03b8 cos\u03b8
\nSolution:
\nDividing by cos2<\/sup> \u03b8 we get
\n\u21d2 sec2<\/sup> \u03b8 + tan2<\/sup> \u03b8 = 3 tan\u03b8
\n\u21d2 1 + 2 tan2<\/sup> \u03b8 = 3 tan\u03b8
\n\u21d2 2tan2<\/sup> \u03b8 – 3tan \u03b8 +1 = \u03b8
\n\u21d2 2 tan2<\/sup> \u03b8 –\u00a0 2tan \u03b8 –\u00a0 tan\u03b8 + 1 = \u03b8
\n\u21d2 (tan\u03b8 –\u00a0 1)(2tan \u03b8 – 1) = \u03b8
\n\u21d2 tan\u03b8 = 1 or tan\u03b8 = \\(\\frac{1}{2}\\)
\nIf tan \u03b8 = 1 then principal solution is \u03b1 =\\( \\frac{\\pi}{4}\\)
\nGeneral solution is \u03b8 = n\u03c0 + \\(\\frac{\\pi}{4}\\)\u00a0 n \u2208 Z
\nLet a be the principal solution of tan \u03b8 = \\(\\frac{1}{2}\\)
\nThen the general solution is
\n\u03b8 = n\u03c0+\u03b1,n\u2208 Z<\/p>\n

Question 3.
\nSolve \\(\\sqrt{2}(\\sin x+\\cos x)=\\sqrt{3}\\)
\nSolution:
\n\"TS<\/p>\n

Question 4.
\nSolve tan\u03b8+3 cot\u03b8 = 5 sec\u03b8
\nSolution:
\nThe given equation is
\ntan\u03b8 + 3 cot\u03b8 = 5 sec\u03b8
\n\u21d2 \\(\\frac{\\sin \\theta}{\\cos \\theta}+\\frac{3 \\cos \\theta}{\\sin \\theta}=\\frac{5}{\\cos \\theta}\\)
\n\u21d2\u00a0 sin2<\/sup> \u03b8 + 3cos2<\/sup>\u03b8= 5 sin\u03b8
\n\u21d2\u00a0 sin2<\/sup> \u03b8 + 3(1 – sin2<\/sup> \u03b8) = 5 sin\u03b8
\n\u21d2 sin2<\/sup> \u03b8 + 3 – 3 sin2<\/sup> \u03b8 = 5 sin\u03b8
\n\u21d2 -2 sin \u03b8 – 5 sin\u03b8 + 3 = 0
\n\u21d2 2 sin \u03b8+ 5 sin\u03b8 – 3 = \u03b8
\n\u21d2 2 sin \u03b8 + 6 sin\u03b8 – sin\u03b8 – 3 = \u03b8
\n\u21d2 2 sin \u03b8(sin \u03b8 + 3) –\u00a0 1(sin\u03b8 + 3) = \u03b8
\n\u21d2 (2sin\u03b8 – 1)(sin\u03b8 + 3)= \u03b8
\n\u21d2 sin \u03b8 = \\(\\frac{1}{2}\\) or sin\u03b8 = – 3.
\nsin\u03b8 =-3 is not admissible but for sin\u03b8 = \\(\\frac{1}{2}\\) general solution is
\n\u03b8=n\u03c0+(-n)n<\/sup> ,n\u2208Z
\nSince the principal solution is \u03b1 = \\(\\frac{1}{2}\\)<\/p>\n

\"TS<\/p>\n

Question 5.
\nIf acos2<\/sup>\u03b8 + bsin2<\/sup>\u03b8 =c has \u03b81<\/sub>,\u03b82<\/sub> as its solutions then show that tan \u03b81<\/sub> + tan \u03b81<\/sub> = \\(\\frac{2 b}{c+a}\\)
\nSolution:
\nGiven equation is acos2\u03b8 + bsin2\u03b8 = c
\n\"TS
\nThis is a quadratic equation in an \u03b8.
\nGiven \u03b81<\/sub>,\u03b82<\/sub> \u00a0are the solutions of \u03b8.
\ntan \u03b81<\/sub>, tan\u03b82 <\/sub>are the roots of equation (1)
\n\u2234 Sum of the roots tan \u03b81<\/sub> + tan \u03b82<\/sub> = \\(\\frac{2 b}{a+c}\\) and product of the roots
\n\"TS<\/p>\n

Question 6.
\nFind all values of x in (-\u03c0, \u03c0) satisfying the equation g1<\/sup>+cosx+cos2<\/sup> x +……………….. = 43<\/sup>
\nSolution:
\n\"TS<\/p>\n

Question 7.
\nSolve sin x = \\(\\frac{1}{\\sqrt{2}}\\)
\nSolution:
\nPrincipal solution is \u03b1 = \\(\\frac{\\pi}{4}\\)
\nGeneral solution is
\n\\(\\mathrm{x}=\\mathrm{n} \\pi+(-1)^{\\mathrm{n}} \\frac{\\pi}{4}, \\mathrm{n} \\in \\mathrm{Z}\\)<\/p>\n

\"TS<\/p>\n

Question 8.
\nSolve sin 2\u03b8 \\(=\\frac{\\sqrt{5}-1}{4}\\)
\nSolution:
\nThe principal solution is \u03b1 = 18\u00b0
\n\u2234 General solution is
\n\"TS
\nQuestion 9.
\nSolve tan2 <\/sup>\u03b8 = 3.
\nSolution:
\n\"TS
\nQuestion 10.
\nSolve 3cosec x = 4sinx.
\nSolution:
\nGiven 3 cosec x = 4 sin x
\n\"TS<\/p>\n

Question 11.
\nIf x is acute and sin(x+10\u00b0) = cos(3x – 68\u00b0) find x in degrees.
\nSolution:
\n\"TS
\nWhen k = 0 we get x = 37\u00b0
\nIf we take k = 1, 2 the value of x is not acute.
\nHence the value of x is 37\u00b0.<\/p>\n

\"TS<\/p>\n

Question 12.
\nSolve cos 3\u03b8 = sin 2\u03b8.
\nSolution:
\n\"TS<\/p>\n

Question 13.
\nSolve 7 sin2<\/sup>\u03b8+3cos2<\/sup>\u00a0\u03b8= 4.
\nSolution:
\nGiven 7 sin2<\/sup>\u03b8+3cos2<\/sup>\u00a0\u03b8= 4.
\n\u21d2\u00a07 sin 2\u03b8 + 3(1 -sin2<\/sup> \u03b8) = 4
\n\u21d2\u00a0 4 sin 2\u03b8 = 1
\n\u21d2 sin\u03b8 = \u00b1 \\(\\frac{1}{2}\\)
\nPrincipal solutions are \\(\\alpha=\\pm \\frac{\\pi}{6}\\) and general solution is given by general solution is given by
\n\u03b8 = n\u03c0 \u00b1 ,\\(\\alpha=\\pm \\frac{\\pi}{6}\\),n\u2208Z<\/p>\n

Question 14.
\nFind general solution of \u03b8 which satisfies both the equations sin\u03b8 = \\(-\\frac{1}{2}\\) and cos\u03b8 = \\(-\\frac{\\sqrt{3}}{2}\\)
\nSolution:
\n\"TS<\/p>\n

Question 15.
\nSolve 4 sin x. sin 2x sin 4x = sin 3x
\nSolution:
\nGiven sin 3x = 4 sinx sin 2x sin 4x
\n= 2sinx (2sin2x – sin4x)
\n\u21d2 2 sin x (cos 2x – cos 6x)
\n\u21d2 sin3x = 2cos 2x sinx – 2 cos 6x sin x
\n\u21d2 sin 3x = sin 3x – sin x – 2 cos 6xsinx
\n\u21d2 2cos 6x sinx + sinx= 0
\n\u21d2 sinx(2 cos 6x + 1)=0
\n\u21d2 sinx = 0 or cos 6x \\(-\\frac{1}{2}\\)<\/p>\n

Case (i): sin x = 0,\u00a0 x= n\u03c0, n\u2208Z is the general solution.
\nCase (ii) : cos 6x = \\(-\\frac{1}{2}\\)
\nPrincipal solution is \u03b1 \\( =\\frac{2 \\pi}{3}\\)
\n\u2234 General solution is 6x = 2\u03c0r \u00b1\u00a0\\(\\frac{2 \\pi}{3}\\)
\n\\(x=\\frac{n \\pi}{3} \\pm \\frac{\\pi}{9}, n \\in Z\\)<\/p>\n

\"TS<\/p>\n

Question 16.
\nIf 0<\u03b8<\u03c0 solve cos\u03b8 cos2\u03b8 cos3\u03b8 = \\(\\frac{1}{4}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 18.
\nSolve sin2x – cos2x = sinx – cosx
\nSolution:
\nThe given equation can be written as
\nsin 2x – sin x = cos 2x – cos x
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 19.
\nSolve cos 3x + cos 2x = sin + sin = \\(\\sin \\frac{3 x}{2}+\\sin \\frac{x}{2} 0 \\leq x \\leq 2 \\pi\\)
\nSolution:
\nGiven cos 3x + cos 2x = sin + sin = \\(\\sin \\frac{3 x}{2}+\\sin \\frac{x}{2} 0 \\leq x \\leq 2 \\pi\\)
\n\"TS
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Inter 1st Year Maths 1A Important Questions Chapter 7 Trigonometric Equations to help strengthen their preparations for exams. TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions Question 1. Solve 2cos2\u03b8 – sin \u03b8+1 = 0 Solution: Question 2. Solve 1 +sin2\u03b8 = 3 sin\u03b8 cos\u03b8 Solution: Dividing by cos2 … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10103"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=10103"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10103\/revisions"}],"predecessor-version":[{"id":10313,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10103\/revisions\/10313"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=10103"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=10103"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=10103"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}