Posted by Help on Sunday, April 1, 2012 at 1:13am.
Solve the equation 2sin(2x) = cos x.
and
Solve the equation
cos^2(2x)–sin^2(2x)= (√3 / 2).

Trig  Reiny, Sunday, April 1, 2012 at 8:01am
1st one:
2sin(2x) = cosx
4sinxcosx  cosx = 0
cosx(4sinx  1) = 0
cosx = 0 or sinx = 1/4
for cosx = 0
x = 90° or 270° OR x = π/2 or 3π/2 radians
for sinx=1/4
x = appr 14.5° or 165.5° OR appr .253 or 2.89 radians

Trig  Reiny, Sunday, April 1, 2012 at 8:10am
2nd:
cos^2(2x)–sin^2(2x)= (√3 / 2)
cos (4x) = √3/2 , using cos (2A) = cos^2 A  sin^2 A
I know cos 30° = cos 330° = √3/2
so 4x = 30° or 4x = 330°
x = 7.5° or x = 82.5°
the period of cos 4x is 360/4 ° = 90°
so adding or subtracting multiples of 90 to any of the above solutions, will yield a new solution
e.g. 7.5 + 90 = 97.5°
97.5+90 = 187.5° etc
check for x = 187.5°
LS = cos^2 375°  sin^2 375°
= .9330...  .0669...
= .866025...
RS = √3/2 = .866025...
= LS
(set your calculator to Radians using the DRG key, and get the equivalent answers in radians)
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