Question

The general solution of $\cos x+\sin x=\cos 2x+\sin 2x$ is

• A. $\left\{2n\pi \right\}\cup \left\{\right(2n+1\left)\frac{\pi }{6}\right\},n\in Z$
• B. $\left\{2n\pi \right\}\cup \left\{\right(4n+1\left)\frac{\pi }{6}\right\},n\in Z$
• C. $\left\{\right(2n+1\left)\frac{\pi }{2}\right\}\cap \left\{\right(2n+1\left)\frac{\pi }{6}\right\},n\in Z$
• D. $\left\{\right(2n+1\left)\frac{\pi }{2}\right\}\cup \left\{\right(4n+1\left)\frac{\pi }{6}\right\},n\in Z$

Answer 1

The correct option is B $\left\{2n\pi \right\}\cup \left\{\right(4n+1\left)\frac{\pi }{6}\right\},n\in Z$

$\cos x+\sin x=\cos 2x+\sin 2x\Rightarrow \cos x-\cos 2x-\sin 2x+\sin x=0\Rightarrow (\cos x-\cos 2x)-(\sin 2x-\sin x)=0\Rightarrow 2\sin \frac{3x}{2}\sin \frac{x}{2}-2\cos \frac{3x}{2}\sin \frac{x}{2}=0\Rightarrow \sin \frac{x}{2}(\sin \frac{3x}{2}-\cos \frac{3x}{2})=0\Rightarrow \sin \frac{x}{2}=0\text{or}\sin \frac{3x}{2}-\cos \frac{3x}{2}=0\Rightarrow \frac{x}{2}=n\pi \Rightarrow x=2n\pi ,n\in Z\sin \frac{3x}{2}-\cos \frac{3x}{2}=0\Rightarrow \sin \frac{3x}{2}=\cos \frac{3x}{2}\Rightarrow \tan \frac{3x}{2}=1\Rightarrow \frac{3x}{2}=n\pi +\frac{\pi }{4}\Rightarrow x=\frac{1}{3}(2n\pi +\frac{\pi }{2})\Rightarrow x=(4n+1)\frac{\pi }{6},n\in Z\therefore x=\left\{2n\pi \right\}\cup \left\{\right(4n+1\left)\frac{\pi }{6}\right\},n\in Z$