Trigonemtric equation Level 1

Exercise:

Solve, $1+\cos x+\cos \left( \frac{x}{2} \right)=0$

Solution: we know that, ${{\cos }^{2}}\theta =\frac{1+\cos 2\theta }{2}\Leftrightarrow 2{{\cos }^{2}}\theta =1+\cos 2\theta $

So $2{{\cos }^{2}}\left( \frac{\theta }{2} \right)=1+\cos \theta $ thus $2{{\cos }^{2}}\left( \frac{x}{2} \right)+\cos \left( \frac{x}{2} \right)=0\Leftrightarrow \cos \frac{x}{2}\left( 2\cos \frac{x}{2}+1 \right)=0$

Put $t=\cos \frac{x}{2}\Leftrightarrow t\left( 2t+1 \right)=0\Leftrightarrow t=0\,\,\,or\,\,t=\frac{-1}{2}$

If \(t=0\Leftrightarrow \cos \left( \frac{x}{2} \right)=0=\cos \left( \frac{\pi }{2} \right)\Leftrightarrow \frac{x}{2}=\pm \frac{\pi }{2}+2k\pi \Leftrightarrow x=\pm \pi +2k\pi \)

If $t=\frac{-1}{2}\Leftrightarrow \cos \left( \frac{x}{2} \right)=\frac{-1}{2}=\cos \left( \frac{2\pi }{3} \right)\Leftrightarrow \frac{x}{2}=\pm \frac{2\pi }{3}+2k\pi \Leftrightarrow x=\pm \frac{4\pi }{3}+2k\pi $