Question

$2^{{\sin }^{2}x}+{5.2}^{{\cos }^{2}x}=7$.

• A. $n\pi +\frac{\pi }{2}$.
• B. $n\pi +\frac{3\pi }{2}$.
• C. $n\pi +\frac{4\pi }{2}$.
• D. None of these

Answer 1

Answer: (A)

Given

$2^{{\sin }^{2}x}+{5.2}^{{\cos }^{2}x}=7$

$2^{(1-{\cos }^{2}x)}+{5.2}^{{\cos }^{2}x}=7$

$\frac{2}{2^{\left({\cos }^{2}x\right)}}+{5.2}^{{\cos }^{2}x}=7$

Let $2^{{\cos }^{2}x}=t$

${\cos }^{2}x\in [0,1]$

$2^{{\cos }^{2}x}\in [2^0,2^1]$

$2^{{\cos }^{2}x}\in [1,2]$

$⟹\frac{2}{t}+5t=7$

$⟹2+5t^2=7t$

$⟹5t^2-7t+2=0$

$⟹5t^2-5t-2t+2=0$

$⟹5t(t-1)-2(t-1)=0$

$⟹t=\frac{2}{5},1$

But $t\in [1,2]$

So $t=1$

$⟹2^{{\cos }^{2}x}=2^0$

$⟹{\cos }^{2}x=0$

$⟹x=n\pi \pm \frac{\pi }{2}$