Homework Help: Trigonometry

Posted by Savannah on Monday, January 16, 2012 at 3:51pm.

2cos^2(x) = 13sinx - 5

How do i solve for x?

I think i have to use the sin^2x + cos^2x = 1, but I'm not sure how to use that. I thought about maybe doing 2(1-sin^2x) = 13sinx - 5 but then i get stuck. Any suggestions?

Trigonometry - Damon, Monday, January 16, 2012 at 3:56pm
2 (1-sin^2 x) = 13 sin 5 - 5

2 - 2 sin^2 x = 13 sin x - 5

2 sin^2 x + 13 sin x - 7 = 0

let z = sin x

2 z^2 + 13 z - 7 = 0

(2z -1)(z+7) = 0

sin x = 1/2 or sin x = -7
-7 is not allowed, beyond range of sin function
so x = 30 degrees
Trigonometry - Savannah, Monday, January 16, 2012 at 4:06pm
Oh that makes sense. That 2sin^2x + 13sinx - 7 can be factored. Thanks!

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