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December 22, 2014

December 22, 2014

Posted by **Bryan** on Sunday, December 9, 2007 at 4:33pm.

8cos^2(X)sin^2(X) + 2cos^2(X) - 3 = 0

There are 8 solutions.

If somebody could show me how to do it and not give me the answers, that would be great.

- Math (Calc) -
**Count Iblis**, Sunday, December 9, 2007 at 5:05pmSubstitute sin^2(x) = 1-cos^2(x) and expand the expression in powers of

cos(x), You get a quadratic equation in cos^2(x). So, if you put cos^2(x) = y, the equation is quadratic in y.

- Math (Calc) -
**Michael**, Sunday, December 9, 2007 at 5:08pmFirst, solve the #1 Identity for sin^2(x) and plug it into the equation so that we have only cosines in the problem. The #1 Identity if sin^2(x) + cos^2(x) = 1.

Distribute.

Factor as you would if the cosine were simply x. Solve for cos(x).

Then, refer to the unit circle (which should be in your head) to solve for x. Cosine corresponds to the x-coordinate.

- Math (Calc) -
**Bryan**, Sunday, December 9, 2007 at 5:21pmI only get 4 answers though. I'm supposed to get 8

- Math (Calc) -
**Count Iblis**, Sunday, December 9, 2007 at 5:29pmIf you put cos(x) = c, the equation becomes:

8c^4 - 10 c^2 + 3 = 0

(4 c^2 - 3)(2c^2 - 1) = 0

c = ±1/2 sqrt(3)

c = ±1/2 sqrt(2)

And you see that:

x = ±pi/6 + n pi

x = ±pi/4 + n pi

There are thus 8 solutions per interval of 2 pi.

- Math (Calc) -
**Bryan**, Sunday, December 9, 2007 at 5:30pmya sorry, I got it confused. Thanks for all the help.

- Math (Calc) -

- Math (Calc) -
- Math (Calc) -
**Damon**, Sunday, December 9, 2007 at 5:31pmif you have something like cos^2 x = 1/16

I am not picking what you actually have of course

then cos x = 1/4 OR cos x = -1/4

now cos 4 = 1/4 at 2 places on the unit circle (first quadrant and fourth)

and

cos x = -1/4 at two places on the circle (second quadrant and third)

that gives you four answers

The other solution for cos^2 gives you four more

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