Question

The number of solutions of the equation $\sin \left(2x\right)+\cos \left(4x\right)=2$ is

• A. $0$
• B. $1$
• C. $2$
• D. $\infty$

Answer 1

The correct option is A

$0$

Explanation for correct option

Given, $\sin \left(2x\right)+\cos \left(4x\right)=2$

$$\begin{gathered} \Rightarrow \sin \left(2x\right)+1-2{\sin }^2\left(2x\right)=2\left[\because \cos \left(2\theta \right)={\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)\&{\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1\right] \\ \Rightarrow \sin \left(2x\right)-2{\sin }^2\left(2x\right)-1=0 \\ \Rightarrow 2{\sin }^2\left(2x\right)-\sin \left(2x\right)+1=0 \\ \therefore \sin \left(2x\right)=\frac{1\pm \sqrt{1-8}}{2·2}=\frac{1\pm i\sqrt{7}}{4}\left[\because i=\sqrt{-1}\right] \end{gathered}$$

Therefore, there are no solution for $x$

Hence, the correct option is $\left(\mathrm{A}\right)$.