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July 23, 2014

July 23, 2014

Posted by **Help** on Sunday, April 1, 2012 at 1:13am.

and

Solve the equation

cos^2(2x)–sin^2(2x)= (√3 / 2).

- Trig -
**Reiny**, Sunday, April 1, 2012 at 8:01am1st 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 8:10am2nd:

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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