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JEE Main 2024
Trigonometric Equations
Trigonometric Equations
Medium

Question

Let S={θ(0,2π):7cos2θ3sin2θ2cos22θ=2}S=\left\{\theta \in(0,2 \pi): 7 \cos ^{2} \theta-3 \sin ^{2} \theta-2 \cos ^{2} 2 \theta=2\right\}. Then, the sum of roots of all the equations x22(tan2θ+cot2θ)x+6sin2θ=0,θSx^{2}-2\left(\tan ^{2} \theta+\cot ^{2} \theta\right) x+6 \sin ^{2} \theta=0, \theta \in S, is __________.

Answer: 7

Solution

7cos2θ3sin2θ2cos22θ=27{\cos ^2}\theta - 3{\sin ^2}\theta - 2{\cos ^2}2\theta = 2 4(1+cos2θ2)+3cos2θ2cos22θ=2 \Rightarrow 4\left( {{{1 + \cos 2\theta } \over 2}} \right) + 3\cos 2\theta - 2{\cos ^2}2\theta = 2 2+5cos2θ2cos22θ=2 \Rightarrow 2 + 5{\cos ^2}\theta - 2{\cos ^2}2\theta = 2 cos2θ=0 \Rightarrow \cos 2\theta = 0 or 52{5 \over 2}(rejected) cos2θ=0=1tan2θ1+tan2θtan2θ=1 \Rightarrow \cos 2\theta = 0 = {{1 - {{\tan }^2}\theta } \over {1 + {{\tan }^2}\theta }} \Rightarrow {\tan ^2}\theta = 1 \therefore Sum of roots =2(tan2θ+cot2θ)=2×2=4 = 2({\tan ^2}\theta + {\cot ^2}\theta ) = 2 \times 2 = 4 But as tanθ=±1\tan \theta = \, \pm \,1 for π4,3π4,5π4,7π4{\pi \over 4},{{3\pi } \over 4},\,{{5\pi } \over 4},\,{{7\pi } \over 4} in the interval (0,2π)(0,\,2\pi ) \therefore Four equations will be formed Hence sum of roots of all the equations =4×4=16. = 4 \times 4 = 16.

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