Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
Find the general solution of the equation $5\cos^2\theta+7\sin^2\theta-6=0$
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Answer
$5\cos^2\theta+7\sin^2\theta-6=0$ $\Rightarrow5\cos^2\theta+7(1-\cos^2\theta)-6=0$$[\because\cos^2\alpha+\sin^2\alpha=1]$ $\Rightarrow5\cos^2\theta+7-7\cos^2\theta-6=0\Rightarrow-2\cos^2\theta+1=0$ $\Rightarrow2\cos^2\theta=1\Rightarrow\cos^2\theta=\frac{1}{2}$ $\Rightarrow\cos^2\theta=\cos^2\frac{\pi}{4}$ $\Rightarrow\frac{1+\cos2\theta}{2}=\frac{1+\cos\frac{\pi}{2}}{2}\Rightarrow\cos2\theta=\cos\frac{\pi}{2}$ $\Rightarrow2\theta=2\text{n}\pi\pm\frac{\pi}{2}$$[\because\cos\theta=\cos\alpha\Rightarrow\theta=2\text{n}\pi\pm\alpha,\text{n}\in\text{z}]$ $\therefore\theta=\text{n}\pi\pm\frac{\pi}{4}$ Hence, the general solution of $\theta=\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{Z}.$
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