How are these solved?
1. cosx+1=0; x ∈ R
2. 4sin^2 x  1 = 0; x ∈ R
3. 2cos^2 x + 3cosx + 1 =0;0 ≤ x < 2π
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1 answer

cosx+1 = 0
cosx = 1
4sin^2x1 = 0
sinx = ±1/2
2cos^2x+3cosx+1 = 0
(2cosx+1)(cosx+1) = 0
cosx = 1/2 or 1
Now find angles in the proper quadrants for those solutions.
And don't forget your Algebra I now that you're doing trig! 👍
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answered by Steve
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