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

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Answer 1 · 2018-03-06T04:12:50

The solutions are $x=sin^-1(-1/3), pi-sin^-1(-1/3)$ .

Use the Pythagorean identity to derive a slightly different one:

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

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

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

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

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

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

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

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

$3(sinx)^2-5sinx-2=0$

Replace $sinx$ with $u$ :

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

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

$u=-1/3, 2$

Substitute $sinx$ back in for $u$ :

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

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

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