Posted by **Sandy** on Sunday, June 6, 2010 at 11:19pm.

solve for x if x is in radians.

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

- trig -
**Reiny**, Sunday, June 6, 2010 at 11:24pm
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 -
**Sandy**, Sunday, June 6, 2010 at 11:27pm
no, its not squared. it is (cosx)(2)

- trig -
**Reiny**, Sunday, June 6, 2010 at 11:42pm
(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 -
**Sandy**, Sunday, June 6, 2010 at 11:49pm
okay i think that is right. thank you

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