Question

The number of solutions of ${16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10$ in the interval $x\in [0,\pi ]$ is

• A. $2$
• B. $8$
• C. $4$
• D. $0$

Answer 1

The correct option is C $4$

${16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10\Rightarrow {16}^{{\sin }^{2}x}+{16}^{1-{\sin }^{2}x}=10\Rightarrow {16}^{{\sin }^{2}x}+\frac{16}{{16}^{{\sin }^{2}x}}=10$
Assuming $y={16}^{{\sin }^{2}x}$
$\Rightarrow y+\frac{16}{y}=10\Rightarrow y^2-10y+16=0\Rightarrow (y-8)(y-2)=0\Rightarrow y=2,8\Rightarrow {16}^{{\sin }^{2}x}=2,8$
Taking $\log$ on both sides,
${\log }_2{16}^{{\sin }^{2}x}={\log }_{2}2,{\log }_{2}8\Rightarrow {\sin }^{2}x\times {\log }_{2}16=1,3\Rightarrow {\sin }^{2}x=\frac{1}{4},\frac{3}{4}\Rightarrow \sin x=\frac{1}{2},\frac{\sqrt{3}}{2}(\because \sin x\ge 0\forall x\in [0,\pi \left]\right)\therefore x=\frac{\pi }{6},\frac{5\pi }{6},\frac{\pi }{3},\frac{2\pi }{3}$