Question

If $4co{s}^2\theta +15cos\theta -4=0$, find the value of $(lo{g}_{2}cos\theta {)}^{-5}$:

• A. $-1$
• B. $-\frac{1}{32}$
• C. $1$
• D. none of these

Answer 1

Answer: (B)

$-\frac{1}{32}$

Given $4{\cos }^2\theta +15\cos \theta -4=0$

$⟹4{\cos }^2\theta +16\cos \theta -\cos \theta -4=0$

$⟹4\cos \theta (\cos \theta +4)-1(\cos \theta +4)=0$

$⟹(4\cos \theta -1)(\cos \theta +4)=0$

$⟹\cos \theta =\frac{1}{4},-4⟹\cos \theta =\frac{1}{4}$

So $({\log }_2\cos \theta {)}^{-5}=({\log }_2{2}^{-2}{)}^{-5}=(-2{)}^{-5}=-\frac{1}{32}$