Question

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

• A. $4$
• B. $8$
• C. $12$
• D. $16$

Answer 1

The correct option is C $12$

Given that,

${16}^{si{n}^{2}x}+{16}^{co{s}^{2}x}=10$ ....$\left(1\right)$

Take ${16}^{si{n}^{2}x}=t$

then, ${16}^{co{s}^{2}x}={16}^{1-si{n}^{2}x}=\frac{16}{{16}^{si{n}^{2}x}}=\frac{16}{t}$

Therefore, ${16}^{co{s}^{2}x}=\frac{16}{t}$

Substituting these values in $\left(1\right)$ we get,

$\Rightarrow t+\frac{16}{t}=10$

$\Rightarrow t^2+16=10t$

$\Rightarrow t^2-10t+16=0$

$\Rightarrow t^2-2t-8t+16=0$

$\Rightarrow t(t-2)-8(t-2)=0$

$\Rightarrow (t-2)(t-8)=0$

$\Rightarrow t=2,t=8$

$\Rightarrow {16}^{si{n}^{2}x}=2or8$

$\Rightarrow {16}^{si{n}^{2}x}={16}^{\frac{1}{4}}or{16}^{\frac{3}{4}}$

$\Rightarrow si{n}^{2}x=\frac{1}{4}or\frac{3}{4}$

$\Rightarrow sinx=\pm \frac{1}{2},\pm \frac{\sqrt{3}}{2}$

Now in $[0,3\pi ]$

$sinx=\frac{1}{2},thenx=\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}$

$sinx=\frac{-1}{2},thenx=\frac{7\pi }{6},\frac{11\pi }{6}$

$sinx=\frac{\sqrt{3}}{2},thenx=\frac{\pi }{3},\frac{2\pi }{3},\frac{7\pi }{3},\frac{8\pi }{3}$

$sinx=\frac{-\sqrt{3}}{2},thenx=\frac{4\pi }{3},\frac{5\pi }{3}$

Hence, there will be $12$ solutions in all