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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^

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?


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Textbook answers:

0, 180, 360
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0 0
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ReportAnswer #1
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
answered by: JimCarrey2JohnnyDepp
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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^
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