Question

Solve the equation 2cos^2x = sin2x , giving your answer in terms of π. 0

Answer 1

Given is
`2*cos^(2) (x)=sin(2x)`

We can use following formulas to solve this problem :-

1.
`(sin\;\theta)/(cos\;\theta) = tan\;\theta`

2. sin(2∅) = 2·sin(∅)·cos(∅)

Solving the given equation :-

`2*cos^(2) (x)=sin(2x) \\\\2*cos^(2) (x)=2*sin(x)*cos(x) \\\\2*cos^(2) (x)-2*sin(x)*cos(x)=0 \\\\2*cos^(2)(x)*(1-(sin(x))/(cos(x))) =0 \\\\2*cos^(2)(x)*(1-tan(x)) =0 \\\\cos^(2)(x) = 0 \;or\; (1-tan(x)) =0 \\\\cos(x) = 0 \;or\; tan(x) =1 \\\\x = cos^(-1)(0) \;or\; x=tan^(-1)(1) \\\\x=(\pi)/(2) \;or\; x=(\pi)/(4)`

Hence, final answer is
`x=(\pi)/(2) \;or\; x=(\pi)/(4)`.

Answer 2

2cos^2x = sin2x 2cos^2x = 2sin x cos x (2sin x cos x) / (2cos^2x) = 1 sin x / cos x = 1 tan x = 1 x = π/4