1. sin identities

Solve equation for values between o and 2 pi

2cos2A + 1 + sinA = 0

2. use the identity:
$\cos{2\theta} = 1 - 2\sin^{2}{\theta}$

Then try to rearrange the equation as necessary for factoring. If need be, use quadratic formula:
$x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$

EDIT: You may also use $\cos^2{\theta} + \sin^2{\theta} = 1$

3. Check this out

Originally Posted by gracey
Solve equation for values between o and 2 pi

2cos2A + 1 + sinA = 0
cos2A=1-2(sinA)^2 therfor
the equation becomes
2(1-2(sinA)^2)+1+sinA=0
2-4(sinA)^2+1+sinA=0
4(sinA)^2-sinA-3=0
this is a quadratic equation.solving this we get
sinA=1 or sinA=-3/4
1)sinA=1
A=pi/2
2)sinA=-3/4
let sinB=3/4(NOT -3/4)
SO B=arcsin3/4
so
A=pi+B,2pi-B

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solve 2cos2a=1 sina

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