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

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Answer 1 · 2016-07-24T03:51:16

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

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

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

Now substituting into our equation:

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

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

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

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

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

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

$sin x = \frac{1}{2} and sin x = - 1$ $sinx = 1/2 and sinx = -1$

$x = \frac{π}{6}, \frac{5 π}{6}, \frac{3 π}{2}$ $x = pi/6, (5pi)/6, (3pi)/2$

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

Hopefully this helps!

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