Question

If $0<\theta ,\varphi <\pi$ and $8\cos \theta \cos \varphi \cos (\theta +\varphi )+1=0$, find $\theta$ and $\varphi$

• A. $\theta =\varphi =\frac{\pi }{6}or\frac{\pi }{3}$
• B. $\theta =\varphi =\frac{\pi }{6}or\frac{2\pi }{3}$
• C. $\theta =\varphi =\frac{\pi }{4}or\frac{\pi }{2}$
• D. $\theta =\varphi =\frac{\pi }{3}or\frac{2\pi }{3}$

Answer 1

Answer: (D)

$\theta =\varphi =\frac{\pi }{3}or\frac{2\pi }{3}$

$\cos \left(\theta \right).\cos \varphi .\cos (\theta +\varphi )=\frac{-1}{8}$
Or
$\cos \left(\theta \right).\cos \varphi .\cos (\theta +\varphi )=\frac{-1}{2^3}$.
Considering
$\cos \left(\theta \right)=\cos \left(\varphi \right)=\cos (\theta +\varphi )=\frac{-1}{2}$.
Hence
$\theta =\varphi ={120}^0=\frac{2\pi }{3}$.
And
$\cos \left(\theta \right).\cos \varphi .\cos (\theta +\varphi )=\frac{1}{2}.\frac{1}{2}.\frac{-1}{2}$.
We get
$\cos \left(\theta \right)=\cos \left(\varphi \right)=\frac{1}{2}$ and $\theta +\varphi =\frac{-1}{2}$.
Hence
$\theta =\varphi =\frac{\pi }{3}$.