Trigonometry question... 2cos²x - 1 = 0 for 0°<=x>=360°?
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
Update:^^cos no sin sorry
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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.
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)
cos^2x = 1/2
cosx = +-sqrt(2)2
x = +-pi/4,+-3pi/4
cos^2x is the same as (cosx)^2.