Use double angle identities to solve trig equations

Question

1. sin 2x - sin x = 0
2. sin 2x sin x = cos x
3. cos 2x - cos x = 0
4. 3 sin^2 x + cos 2x -2 = 0

Answer 1

1. x=0 or pi + 2npi where n = any integer. sin2x=sinx when 2x=x x=0

Use the double angle formula: substitute 2sinxcosx for sin2x

sin2x -sinx = 0

2sinxcosx-sinx=0

factor out sinx to get sinx(2cosx-1) = 0

set each factor = 0

sinx=0

x = inverse sine of x = 0 or pi +2npi where n= any integer

2cosx-1 = 0

2cosx = 1

cosx= 1/2

x = the angle whose cosine is 1/2 = 60 degrees = pi/3 and -60 degrees = -pi/3

both answers include + 2npi where n = any integer

2 sin2xsinx = cosx

2sinxcosxsinx = cosx

divide by cosx

2sinxsinx = 1

sin²x = 1

square root of both sides

sinx = -1, 1

x = 90 or 270 + n720, or pi/2 or 3pi/2 + 2npi where n= any integer

3 is very similar to 1, just with cosines instead of sines

cos2x = cosx

substitute the double angle formula cos2x = 2cos²x-1,

2cos²x-1 = cosx

2cos²x - cosx -1 = 0

factor

(2cosx+1)(cosx-1) = 0

now set each factor = 0

2cosx=1 and cosx=1

cosx=1/2 and 1

x = +60 degrees, 0, each +720n where n=any integer

or = plus or minus +pi/3, 0 each +2npi where n=any integer

4 3sin²x + cos2x -2 = 0, substitute the double angle formula

3sin²x + (1-2sin²x) - 2 = 0, simplify

sin²x -1 = 0

sin²x = 1

sinx = -1, 1

x = 90, 270 degrees = pi/2, 3pi/2 each with +2npi

It helps to check each answer, plugging it back into the original equation

3(1) + cos180 -2 = 0

cos180 = -1

x=90 works

3(-1)² + cos540 -2 = 0

3 + cos180 -2 = 0 cos540=cos(540-360) = cos180

cos180 = -1

x=180 works too