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Solve each of the following for "x" where 0

Mathematics
Tutor: None Selected Time limit: 1 Day

cos^2x-cosx-2=0

4sin^2x+2=3

Mar 20th, 2015

Factor the first equation: cos^2 x - cos x - 2 = (cos x - 2)(cos x + 1) = 0.

Since -1 <= cos x <= 1, the first factor is not 0, and cos x + 1 = 0.

From cos x = - 1 we get x = pi + 2*pi*n (n is any integer).

The second equation can be solved as follows: 4 sin^2 x + 2 = 3; 4 sin^2 x = 1; sin^2 x = 1/4; sinx = +/- 1/2,

and x = +/- pi/6 + pi*n (n is any integer).

Mar 20th, 2015

If you add the restriction 0 < x < 2 pi,

then the first equation has only one solution, x = pi;

the second equation has four solutions: x = pi/6, x=5*pi/6, x = 7*pi/6, and x = 11*pi/6.

Mar 20th, 2015

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