How do you solve $2cos^2x-sinx-1=0$ ?

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Answer 1 · 2016-07-24T03:51:16

Let's first of all convert everything to $sinx$ .

Consider the pythagorean identity $sin^2x + cos^2x = 1$ . Rearranging:

$cos^2x = 1 - sin^2x$

Now substituting into our equation:

$2(1 - sin^2x) - sinx - 1 = 0$

$2 - 2sin^2x - sinx - 1 =0$

$-2sin^2x - sinx + 1 = 0$

$-2sin^2x - 2sinx + sinx + 1 = 0$

$-2sinx(sinx + 1) + 1(sinx + 1) = 0$

$(-2sinx + 1)(sinx + 1) = 0$

$sinx = 1/2 and sinx = -1$

$x = pi/6, (5pi)/6, (3pi)/2$

Note that these solutions are only in the interval $0 < x < 2pi$ .

Hopefully this helps!

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