SOLUTION: 2sec^2 x-5 sec x+2=0 cos 2x-2 sinx- cos^2 x=-3

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Question 1161776: 2sec^2 x-5 sec x+2=0
cos 2x-2 sinx- cos^2 x=-3
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


2sec%28x%29%5E2-5sec%28x%29%2B2+=+0

Factor like a quadratic, treating "sec(x)" as the variable.

%282sec%28x%29-1%29%28sec%28x%29-2%29+=+0
sec%28x%29+=+1%2F2%29 OR sec%28x%29+=+2

Take reciprocals to change sec(x) to cos(x).

cos%28x%29+=+2 OR cos%28x%29+=+1%2F2

You can finish from there....

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cos%282x%29-2sin%28x%29-cos%28x%29%5E2+=+-3

Substitute for cos(2x); use cos%28x%29%5E2-sin%28x%29%5E2, since the cosines will cancel, leaving a quadratic in sin(x).

%28cos%28x%29%5E2-sin%28x%29%5E2%29-2sin%28x%29-cos%28x%29%5E2+=+-3
-sin%28x%29%5E2-2sin%28x%29+=+-3
sin%28x%29%5E2%2B2sin%28x%29-3+=+0
%28sin%28x%29%2B3%29%28sin%28x%29-1%29+=+0
sin%28x%29+=+-3 OR sin%28x%29+=+1

Again you can finish....