Question

$y=f\left(x\right)={\log }_2(2-{\log }_{\sqrt{2}}(16{\sin }^{2}x+1))$ Then find $y$ when $x=0$.

Answer 1

$y=f\left(x\right)={\log }_2(2-{\log }_{\sqrt{2}}(16{\sin }^{2}x+1))$

$y=f\left(x\right)={\log }_2(2-{\log }_{\sqrt{2}}(16{\sin }^{2}0+1))$

$y=f\left(x\right)={\log }_2(2-{\log }_{\sqrt{2}}\left(16\right(0)+1))$

$y=f\left(x\right)={\log }_2(2-{\log }_{\sqrt{2}}\left(1\right))$

$y=f\left(x\right)={\log }_2(2-0)$

$y=f\left(x\right)={\log }_{2}2=1$