The number of solutions of the trigonometric equation $1+\cos x \cdot \cos 5 x=\sin ^2 x$ in $[0,2 \pi]$ is

The number of solutions of the trigonometric equation $1+\cos x \cdot \cos 5 x=\sin ^2 x$ in $[0,2 \pi]$ is
  1. 8
  2. 12
  3. 10
  4. 6

Solution

We have, $\begin{array}{ll} & 1+\cos x \cdot \cos 5 x=\sin ^2 x \\ \Rightarrow & 1-\sin ^2 x+\cos x \cdot \cos 5 x=0 \\ \Rightarrow \quad & \cos ^2 x+\cos x \cdot \cos 5 x=0 \\ \Rightarrow \quad & \cos x(\cos x+\cos 5 x)=0 \\ \Rightarrow \quad & \cos x\left[2 \cos \left(\frac{x+5 x}{2}\right) \cos \left(\frac{x-5 x}{2}\right)\right]=0 \\ \Rightarrow \quad & \cos x(2 \cos 3 x \cos 2 x)=0 \\ \Rightarrow \quad & \cos x=0 ; \cos 3 x=0 \text { or } \cos 2 x=0 \\ \Rightarrow \quad & x=\frac{\pi}{2}, \frac{3 \pi}{2} \text { or } \\ & 3 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}, \frac{9 \pi}{2}, \frac{11 \pi}{2} \text { or } \\ & 2 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \\ \Rightarrow \quad & x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \end{array}$ Thus, there are 10 solutions in $[0,2 \pi]$

Asked in: MHT CET Full Test 7

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