How do you solve $3 {cos}^{2} x + 5 sin x - 1 = 0$ $3cos^2x +5sinx - 1=0$ ?

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Answer 1 · 2018-03-06T04:12:50

The solutions are $x = {sin}^{- 1} (- \frac{1}{3}), π - {sin}^{- 1} (- \frac{1}{3})$ $x=sin^-1(-1/3), pi-sin^-1(-1/3)$ .

Use the Pythagorean identity to derive a slightly different one:

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

$\Rightarrow {cos}^{2} x = 1 - {sin}^{2} x$ $=>cos^2x=1-sin^2x$

Use this new identity to replace ${cos}^{2} x$ $cos^2x$ in the actual problem, then solve it like a quadratic:

$3 {cos}^{2} x + 5 sin x - 1 = 0$ $3color(red)(cos^2x)+5sinx-1=0$

$3 (1 - {sin}^{2} x) + 5 sin x - 1 = 0$ $3(color(red)(1-sin^2x))+5sinx-1=0$

$3 - 3 {sin}^{2} x + 5 sin x - 1 = 0$ $3-3sin^2x+5sinx-1=0$

$- 3 {sin}^{2} x + 5 sin x + 2 = 0$ $-3sin^2x+5sinx+2=0$

$3 {sin}^{2} x - 5 sin x - 2 = 0$ $3sin^2x-5sinx-2=0$

$3 {(sin x)}^{2} - 5 sin x - 2 = 0$ $3(sinx)^2-5sinx-2=0$

Replace $sin x$ $sinx$ with $u$ $u$ :

$3 u^{2} - 5 u - 2 = 0$ $3u^2-5u-2=0$

$(3 u + 1) (u - 2) = 0$ $(3u+1)(u-2)=0$

$u = - \frac{1}{3}, 2$ $u=-1/3, 2$

Substitute $sin x$ $sinx$ back in for $u$ $u$ :

$sinx=-1/3,color(red)cancelcolor(black)(sinx=2)$

$x=sin^-1(-1/3), pi-sin^-1(-1/3)$

How do you solve 3cos^2x +5sinx - 1=03cos2x+5sinx−1=03cos^2x +5sinx - 1=0?