Question

Solve the following equations.
$\sqrt{co{s}^{2}2x+|sin(2x-\frac{3}{2}\pi )|+\frac{1}{4}}=cos\left(\frac{20}{12}\pi \right)$

Answer 1

$\sqrt{co{s}^{2}2x+|sin(2-\frac{3}{2}\pi )|+\frac{1}{4}}=cos\left(\frac{20}{12}\pi \right)-$

$So{l}^n$ we know that root always having positive quantity and also modules.

i.e.$\sqrt{co{s}^2\left(2x\right)+|sin(2x-3/2\pi )|+\frac{1}{4}}=cos\left(\frac{20}{12}\pi \right)...\left(1\right)$

In eq (1) we can see

$\left|sin\right(2x-3/2\pi \left)\right|=|sin2xcos\frac{3}{2}\pi -cos2xsin\frac{3}{2}\pi |$

$\Rightarrow |sin2x\times 0-cos2x\times 1|$

$\Rightarrow |-cos2x|$

Putting the above value in equation (1)

i.e.

$\sqrt{co{s}^2\left(2x\right)+cos2x+\frac{1}{4}}=cos\left(\frac{20}{12}\pi \right)$

Now we use perfect square method

$\therefore \sqrt{co{s}^{2}2x+2\times \frac{1}{2}cos2x+{\left(\frac{1}{2}\right)}^2}=cos\left(\frac{20}{12}\pi \right)(a+b{)}^2=a^2+2ab+b^2$

$\Rightarrow \sqrt{(cos2x+\frac{1}{2}{)}^2}=cos\left(\frac{20}{12}\pi \right)$

$\Rightarrow cos(2x+\frac{1}{2})=cos\left(\frac{5}{3}\pi \right)$

$\therefore 2x+\frac{1}{2}=\frac{5}{3}\pi$

$\Rightarrow 2x=\frac{5}{3}\pi -\frac{1}{2}$

$x=\frac{5}{6}\pi -\frac{1}{4}$