Trigonometry question... 2cos²x - 1 = 0 for 0°<=x>=360°?

Question

Solve 2cos²x - 1 = 0 for 0°<=x>=360°

please could someone help me with this question?
it has got me stuck because cos² rather than just cos.

Thx for the help guys

This is what my approach would be, but the answer in my book says i am wrong.

i would substitue x² for cos²x.

giving me, 2x²-1 = 0
take the 1 over, 2x² = 1
divide by 2, x² = 0.5
sqrt x = 0.7071, or , -0.7071

x = arcsin
x = 45, or ,-45

i would then work out which other sin values equal 45 and -45 between 0° and 360°. but my answer says that 45, and -45 are not any of the answer...

help please? thx

^^cos no sin sorry

Answer 1

cos^2 x = 1/2
cos x = sq rt (1/2)
x = 45, 135, 225, 315
your answers are right -- 45 is one of them, and -45 degrees is the same as 315. There are a couple others too but it looks like you would have foudn them.

Answer 2

2cos²x - 1 = 0

2cos²x = 1

cos²x = 1/2

cos x = +- sqrt(2)/2

If cos(x) = sqrt(2)/2 then

x = +- 45° + 360°n for n=0,1,2,...

x=45 or 315

If cos(x) = -sqrt(2)/2

x = +-135° + 360°n for n=0,1,2,...

x=135 or 225

Unite both cases and get

x in (45,135,225,315)

Answer 3

cos^2x = 1/2
cosx = +-sqrt(2)2
x = +-pi/4,+-3pi/4

Answer 4

cos^2x is the same as (cosx)^2.