Question

Solve the following equations.
$\frac{3}{\sqrt{4cos2x+1}}=\sqrt{2cos2x+2}.$

Answer 1

$\frac{3}{\sqrt{4cos2x+1}}=\sqrt{2cos2x+2}$

$3=\sqrt{8co{s}^{2}2x+10cos2x+2}$

$3=\sqrt{8(1-2si{n}^{2}x{)}^2+10(1-2si{n}^{2}x)+2}$

$[\therefore cos2x=1-2si{n}^{2}x$

squaring both sides

$a=20+32si{n}^{4}x-52si{n}^{2}x$

take $si{n}^{2}x=t$

then, $t^2-\frac{52}{32}t+\frac{11}{32}=0$

$t=\frac{52/32\pm \sqrt{(52/32{)}^2-4\times 11/32}}{2}$

$t=1.375$

OR

$0.25$

$si{n}^{2}x=1.375$

OR

$0.25$

$sinx=\pm 1.17$ OR $\pm 0.5$

$sinx=\pm 1/2$

$x=\pi /6,2\pi /3,7\pi /6,5\pi /3$