Posted by Anonymous on Thursday, November 22, 2007 at 5:35pm.
Solve the following equation for
0 less than and/or equal to "x" less than and/or equal to 360

cos^2x  1 = sin^2x

Attempt:
cos^2x  1  sin^2x = 0
cos^2x  1  (1  cos^2x) = 0
cos^2x  1  1 + cos^2x = 0
2cos^2x  2 = 0
(2cos^2x/2)= (2/2)
cos^2x = 1
cosx = square root of 1
And I can't do anything with this now...what am I doing wrong?

Textbook answers:
0, 180, 360


Math  Solving for Trig Equations  bobpursley, Thursday, November 22, 2007 at 5:40pm
On the fourth/fifth line, you erred. When you add 2 to both sides, the right side is positive, not negative.
cos^2 x=1
cos x= + 1
giving 1, 180,360
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