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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

The number of solutions of the trigonometric equation 1+ x · 5 x=...

The number of solutions of the trigonometric equation 1+cos⁡x⋅cos⁡5x=sin⁡2x1+\cos x \cdot \cos 5 x=\sin ^2 x in [0,2π][0,2 \pi] is

Options

  1. A8
  2. B12
  3. C10
  4. D6

Correct answer

C. 10

Step-by-step solution

We have, 1+cos⁡x⋅cos⁡5x=sin⁡2x⇒1−sin⁡2x+cos⁡x⋅cos⁡5x=0⇒cos⁡2x+cos⁡x⋅cos⁡5x=0⇒cos⁡x(cos⁡x+cos⁡5x)=0⇒cos⁡x[2cos⁡(x+5x2)cos⁡(x−5x2)]=0⇒cos⁡x(2cos⁡3xcos⁡2x)=0⇒cos⁡x=0;cos⁡3x=0 or cos⁡2x=0⇒x=π2,3π2 or 3x=π2,3π2,5π2,7π2,9π2,11π2 or 2x=π2,3π2,5π2,7π2⇒x=π2,3π2,π6,5π6,7π6,11π6,π4,3π4,5π4,7π4\begin{array}{ll}& 1+\cos x \cdot \cos5 x=\sin ^2 x \\ \Rightarrow & 1-\sin ^2 x+\cos x \cdot \cos5 x=0 \\ \Rightarrow \quad & \cos ^2 x+\cos x \cdot \cos5 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 } \cos2 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 [0,2π][0,2 \pi]
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