Question

Solve the equation.

Answer 1

\[\Huge 16^{\sin^2x}+16^{\cos^2x}=1\]
x=?

Answer 2

Write sin^2 x as 1-cos^2 x and simplify.. Just a guess.

Answer 3

\[\LARGE 16^{1-\cos^2x}+16^{\cos^2x}=16 \times \frac{1}{16^{\cos^2x}}+16^{\cos^2x}=1\]

Answer 4

Put 16cos^2x=t..
The resulting equation will be a quadratic.

Answer 5

\[\LARGE 16 \times \frac{1}{t}+t=t^2+16t=t\]

Answer 6

\[\LARGE t^2+15t=0=>t(t+15)=0\]

Answer 7

\[\Huge 16^{\sin^2x}+16^{\cos^2x}=10\]
number of solutions?
a little change in question

Answer 8

0<x<2pi

Answer 9

i start from where you left.
either t=0 or t=-15
\[8*2\cos ^{2}x=0,8\left( 1+\cos 2x \right)=0,1+\cos 2x=0,\cos2x=-1 \]
\[\cos 2x=-1=\cos \left( 2n+1 \right)\pi,n=0,1\]
\[2x=\pi,3\pi,x=\frac{\pi }{ 2 },\frac{ 3 \pi }{ 2 } \]

Answer 10

\[\LARGE 16^{\sin^2x}+\frac{16}{16^{\sin^2x}}=10\]
Let 16^{sin^2x}=t
\[\LARGE y+\frac{16}{y}=10=>y^2-10y+16=0\]
\[\large y^2-10y+16=0=>y^2-8y-2y+16=0=>(y-2)(y-8)=0\]
\[\Huge y=2,8\]
\[\Huge 16\sin^2x=2,16\sin^2x=8\]
\[\Huge \sin x=\frac{1}{2},\frac{-1}{2},\frac{\sqrt{3}}{2},\frac{-\sqrt{3}}{2}\]