Question

Total number of solution of $2^{\cos x}=|\sin x|$ in $[-2\pi ,5\pi ]$ is equal to

• A. $12$
• B. $14$
• C. $16$
• D. $15$

Answer 1

The correct option is B $14$

According to question..........

$\begin{array}{c}\frac{1}{2}\le 2^{\cos x}\le 2|weknow,max\&minvalueofthecosx=1or-1\\ so,\\ when,\\ \cos x=1,\Rightarrow x=0,2\pi ,4\pi \\ \cos x=0\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2}\\ and,\\ \cos x=-1\Rightarrow x=\pi ,3\pi \\ weknowpointsupto5\pi .\\ 0-\pi \Rightarrow 2\\ \pi -2\pi \Rightarrow 2\\ 2\pi -3\pi \Rightarrow 2\\ 3\pi -4\pi \Rightarrow 2\\ 4\pi -5\pi =2\\ Now,wefindoutfrom[-2\pi ,5\pi ]:\\ [-2\pi ,5\pi ]\\ -2\pi -(-\pi )=2\\ -\pi -0\Rightarrow 2\\ 0-\pi \Rightarrow 2\\ \pi -2\pi \Rightarrow 2\\ 2\pi -3\pi \Rightarrow 2\\ 3\pi -4\pi \Rightarrow 2\\ 4\underset{–}{\pi -5\pi =2}\\ sothetotalno.ofsolutionis14,andthecorrectoptionisB.\end{array}$