Asked by Anonymous
                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
--
            
        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
--
Answers
                    Answered by
            bobpursley
            
    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
    
cos^2 x=1
cos x= +- 1
giving 1, 180,360
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