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September 30, 2014

September 30, 2014

Posted by **arka bhaskar** on Sunday, January 29, 2012 at 9:31am.

- trig -
**drwls**, Sunday, January 29, 2012 at 10:17amsin 2x = cos3x = sin(pi/2 - 3x)

Therefore, one solution is

2x = pi/2 - 3x

5x = pi/2

x = pi/10 = 18 degrees

There is a second solution at x = pi/2, but that is just outside the domain of interest.

- trig -
**Damon**, Sunday, January 29, 2012 at 10:25amDo you mean

sin^2 x = cos^3 x

or do you mean

sin(2x) = cos(3x)

?????

If you mean sin(2x) = cos(3x)

then

2 sin x cos x = 4 cos^3 x - 3 cos x

2 sin x = 4 cos^2 x - 3

2 sin x = 4 (1 - sin^2 x) - 3

2 sin x = 4 - 4 sin^2 x - 3

4 sin^2 x + 2 sin x -1 = 0

let z = sin x

4 z^2 + 2 z - 1 = 0

z = [ -2 +/- sqrt (4+16) ] / 8

z = [ -2 +/- 2 sqrt 5 ] /8

or

sin x = (-1 +/- sqrt 5)/4

if x in quadrant 1 then use + root

sin x = (sqrt 5 - 1)/4

- trig -
**Damon**, Sunday, January 29, 2012 at 10:27amDo it the way Dr WLS did it, much simpler and more likely correct.

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