How to solve this trig equation in radians 2sin^2(x) + sin(x) - 1=0

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To solve the equation 2sin^2(theta) + sin(theta) - 1 = 0 for theta in the interval [0, 2pi], substitute sin(theta) with x, transforming the equation into 2x^2 + x - 1 = 0. This quadratic equation can be solved using the quadratic formula or factoring. After finding the values of x, convert them back to angles theta using the inverse sine function. The discussion emphasizes the importance of understanding the transformation and solving for the angles within the specified interval.
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I just got a review sheet in math that contain questions that I don't know how to do from last year. ONe asks to solve for all values of theta of [0,2pi] in radians: 2sin^2(theta)+sin(theta)-1=0

any ideas?
 
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Use
sin\ 2\theta = 2\ sin\ \theta\ cos\ \theta
sin^2\ \theta + cos^2\ \theta = 1
 
that confuses me more. I have no clue what the question is even asking, I think I forgot all my math. Could you give a longer explanation?
 
Let sin\ \theta\ = x

Then your original equation now looks like 2x^2 + x - 1 = 0.

Solve the quadratic equation. Once you do that, the rest should be easy.
 
UrbanXrisis said:
that confuses me more. I have no clue what the question is even asking, I think I forgot all my math. Could you give a longer explanation?

The question is asking for which angles in a 0 to 2pi interval satisfy that equation.

-Cyclovenom
 
forget about what i have said. use recon's suggestion and understand cyclovenom's point.
 
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