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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the general solution of the equation $5\cos^2\theta+7\sin^2\theta-6=0$

$\begin{array}{1 1}(A)\;n\pi\pm \large\frac{\pi}{4}&(B)\;n\pi\pm \large\frac{\pi}{6}\\(C)\;n\pi\pm \large\frac{\pi}{8}&(D)\;2n\pi\pm \large\frac{3\pi}{4}\end{array} $

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  • $(1-\sin^2\theta)=\cos^2\theta$
$5\cos^2\theta+7\sin^2\theta-6=0$
$5(1-\sin^2\theta)+7\sin^2\theta-6=0$
$5-5\sin^2\theta+7\sin^2\theta-6=0$
$2\sin^2\theta-1=0$
$\sin^2\theta=\large\frac{1}{2}$
$\sin \theta=\pm \large\frac{1}{\sqrt 2}$
$\theta=\large\frac{\pi}{4}$ or $\large\frac{5\pi}{4}$
$\therefore$ General solution =$n\pi\pm \large\frac{\pi}{4}$
Hence (A) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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