Question

Question #74920

Answer 1

`2sin^2x+sin2x=0`

`=>2sin^2x+2sinxcosx=0`

`=>2sinx(sinx+cosx)=0`

So `sinx=0=sin0=sinpi=sin2pi`

`=>x=0 or pi or 2pi`

`=>x=0^@ or 180^@ or 360^@`

Again

#=>sinx+cosx=0

`=>sinx=-cosx`

#=>sinx/cosx=-1

`=>tanx=-1`

`=>tanx=-tan(pi/4)=tan(pi-pi/4)=tan(2pi-pi/4)`

So

`x=(3pi)/4 or 135^@ and (7pi)/4 or 315^@`

Answer 2

`0, (3pi)/4, pi, (7pi)/4, 3pi`

Explanation:

`2sin^2 x + 2sin x.cos x = 0`
sin x(sin x + cos x) = 0
a. sin x = 0 --> 3 solution arcs -->
x = 0, `x = pi`, `x = 2pi`
b. sin x + cos x = 0
From trig identity :
`sin x + cos x = sqrt2cos (x - pi/4)`, we get:
`cos (x - pi/4) = 0` --> 2 solutions -->
`x - pi/4 = pi/2` --> `x = pi/2 + pi/4 = (3pi)/4`
`x - pi/4 = (3pi/2)` --> `x = (3pi)/2 + pi/4 = (14pi)/8 = (7pi)/4`
Answers for `(0, 2pi)`:
`0, (3pi)/4, pi, (7pi)/4, 2pi`