Question

Solve the following equations.
$4^{3+2cos2x}-{7.4}^{1+cos2x}-4^{1/2}=0$

Answer 1

$4^{3+2cos2x}-{7.4}^{1+cos2x}-4^{1/2}=0$

$4^{1+2(1+cos2x)}-{7.4}^{1+cos2x}-2=0$

${4.4}^{2(1+cos2x)}-{7.4}^{1+cos2x}-2=0$

$4^{2+cos2x}=t$

$4.t^2-7t-2=0$

$4t^2-8t+2-2=0$

$4t(t-2)+(t-2)=0$

$(t-2)(4t+1)=0$

$t=2,t=\frac{-1}{4}$

$t=\frac{-1}{4}$ (Rejected)

$4^{1+cos2x}=2$

$4^{1+cos2x}=4^{1/2}$

$1+cos2x=1/2$

$cos2x=\frac{-1}{2}$

$2x=\frac{2\pi }{3}$

$x=\frac{\pi }{3}$