# trig

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solve for x if x is in radians.

cos2(x) + 4cos(x) = -3

• trig -

Is that
cos^2 x + 4cosx = -3 ??
if so, then let cosx = y to get
y^2 + 4y + 3 = 0
(y+1)(y+3) = 0
y = -1 or y = -3
so
cosx = -1 ---> x = π
or
cosx = -3 , not possible

so x = π

• trig -

no, its not squared. it is (cosx)(2)

• trig -

(cosx)(2) wouldn't make much sense, then it would be
2cosx and the entire equation would be
6cosx = -3
cosx = -1/2
and x = 120° or 240°

did you mean cos(2x) ?
then you can use the identity
cos(2x) = 2cos^2x - 1

cos((x) + 4cos(x) = -3
2cos^2x - 1 + 4cosx = -3
2cos^2x + 4cosx + 2 = 0
cos^2x + 2cosx + 1 = 0
(cosx + 1)^2 = 0
cosx + 1 = 0
cosx = -1
x = 180° or x = π
(which by coincidence is the same answer to the other interpretation)

• trig -

okay i think that is right. thank you

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