Question

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

Answer 1

Answer and Explanation:

Given : Equation
`2\cos^2x = \sin2x`

To find : Solve the given equation ?

Solution :

Equation
`2\cos^2x = \sin2x`

`2\cos^2x-\sin2x=0`

Open identity,
`\sin 2x=2\sin x\cos x`

`2\cos^2x-2\sin x\cos x=0`

`2\cos x(\cos x-\sin x)=0`

`2\cos x=0,(\cos x-\sin x)=0`

1) When
`2\cos x=0`

`\cos x=0`

General solution -
`x=(\pi)/(2)+n\pi`

From
`(0,2\pi)` the solution is
`((\pi)/(2),0),((3\pi)/(2),0)`

2) When
`(\cos x-\sin x)=0`

`\cos x=\sin x`

`1=(\sin x)/(\cos x)`

`\tan x=1`

General solution -
`x=(\pi)/(4)+n\pi`

From
`(0,2\pi)` the solution is
`((\pi)/(4),0),((5\pi)/(4),0)`

Answer 2

2cos^2x = sin2x
As,
sin2x = 2sinxcosx
2cos^2(x) = 2 sinx cosx
Dividing by 2 on both sides:
cos^2x = sinx cosx
Taking sinx cosx into left side:
cos^2x - sinx cosx = 0

cosx (cosx - sinx) = 0

cosx = 0 ---> x = pi/2 or cosx - sinx = 0

cosx = sinx -----> x = pi/4

so answer is {pi/4 , pi/2}