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March 26, 2015

March 26, 2015

Posted by **Michael** on Tuesday, January 11, 2011 at 8:02pm.

sin2xsinx=cosx

cos2x-cosx=0

- math-trigonometry -
**Reiny**, Tuesday, January 11, 2011 at 8:40pmare we solving?

1st one:

sin2xsinx=cosx

2sinxcosxsinx - cosx = 0

cosx(2sin^2 x - 1) = 0

cosx = 0 or sin^2x = 1/2

if cosx=0,

x = π/2 or x=3π/2 (90° or 270°)

if sin^2x = 1/2

sinx = ±1/2

x = π/4 , 3π/4, 5π/4, or 7π/4 (45°, 135°, 225° or 315°)

for the 2nd

use cos 2x = 2cos^2x - 1 to end up with a quadratic.

give it a try.

it will factor

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