Question

Total number of solutions of ${16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10$ in $x\in [0,3\pi ]$ is equal to

• A. 4
• B. 8
• C. 12
• D. 16

Answer 1

The correct option is C 12

${16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10,$ let $t={16}^{{\sin }^{2}x}$
$\Rightarrow t+\frac{16}{t}=10\Rightarrow t^2-10t+16=0$
$\Rightarrow (t-8)(t-2)=0\Rightarrow t=8,2$
$\Rightarrow {16}^{{\sin }^{2}x}=8,2\Rightarrow {\sin }^{2}x=\frac{3}{4},\frac{1}{4}$
$\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2},\pm \frac{1}{2}$
Thus, there are 12 solutions in $[0,3\pi ].$