Question

Number of solutions to the equation $2^{{\sin }^{2}x}+{5.2}^{{\cos }^{2}x}=7,$ in the interval $[-\pi ,\pi ],$ is

• A. $2$
• B. $4$
• C. $6$
• D. none of these

Answer 1

The correct option is A $2$

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

$2^{{\sin }^{2}x}+{5.2}^{1-{\sin }^{2}x}=7$

Let $2^{{\sin }^{2}x}=k$

$k+5\cdot \frac{2}{k}=7$

$k+\frac{10}{k}=7$

$k^2+10=7h$

$k^2-7k+10=0$

$(k-2)(k-5)=0$

But $2^{{\sin }^{2}x}\le 2^{\prime }=2$

Hence $2^{{\sin }^{2}x}\ne 5$

${\sin }^{2}x=1$

$\Rightarrow x=\frac{\pi }{2},\frac{-\pi }{2}$

Hence there are only two solution in $[-\pi ,\pi ]$ i.e., $\frac{-\pi }{2}$ and $\frac{\pi }{2}$.