How do you solve $2 {(sin x)}^{2} - 5 cos x - 4 = 0$ $2(sinx)^2 - 5cosx - 4 = 0$ in the interval [0, 2pi]?

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How do you solve $2 {(sin x)}^{2} - 5 cos x - 4 = 0$ $2(sinx)^2 - 5cosx - 4 = 0$ in the interval [0, 2pi]?

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Answer 1 · 2016-05-03T09:01:51

$x = \frac{2 π}{3}$ $x=(2pi)/3$ or $\frac{4 π}{3}$ $(4pi)/3$

Explanation:

$2 {(sin x)}^{2} - 5 cos x - 4 = 0$ $2(sinx)^2-5cosx-4=0$

is equivalent to

$2 (1 - {cos}^{2} x) - 5 cos x - 4 = 0$ $2(1-cos^2x)-5cosx-4=0$ or

$2 - 2 {cos}^{2} x - 5 cos x - 4 = 0$ $2-2cos^2x-5cosx-4=0$

or $2 {cos}^{2} x + 5 cos x + 2 = 0$ $2cos^2x+5cosx+2=0$

or using quadratic formula

$cos x = \frac{- 5 \pm \sqrt{5^{2} - 4 \times 2 \times 2}}{2 \times 2} = \frac{- 5 \pm \sqrt{25 - 16}}{4}$ $cosx=(-5+-sqrt(5^2-4xx2xx2))/(2xx2)=(-5+-sqrt(25-16))/4$

or $cos x = \frac{- 5 \pm 3}{4}$ $cosx=(-5+-3)/4$ or $cos x = - 2$ $cosx=-2$ or $cos x = - \frac{2}{4} = - \frac{1}{2}$ $cosx=-2/4=-1/2$

But $cos x$ $cosx$ cannot be $- 2$ $-2$ , hence $cos x = - \frac{1}{2}$ $cosx=-1/2$

or $x = 2 n π \pm \frac{2 π}{3}$ $x=2npi+-(2pi)/3$

And in interval $[0, 2 π]$ $[0,2pi]$ , $x = \frac{2 π}{3}$ $x=(2pi)/3$ or $\frac{4 π}{3}$ $(4pi)/3$

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How do you solve $2 {(sin x)}^{2} - 5 cos x - 4 = 0$ $2(sinx)^2 - 5cosx - 4 = 0$ in the interval [0, 2pi]?

1 Answer

Explanation:

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How do you solve 2(sinx)^2 - 5cosx - 4 = 02(sinx)2−5cosx−4=0 2(sinx)^2 - 5cosx - 4 = 0 in the interval [0, 2pi]?

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How do you solve $2 {(sin x)}^{2} - 5 cos x - 4 = 0$ $2(sinx)^2 - 5cosx - 4 = 0$ in the interval [0, 2pi]?