# Thread: finding x value in trig equation

1. ## finding x value in trig equation

hello can someone please demonstrate how a question like this is solved for the the value of (x) between 0 and 360 degrees.

$\displaystyle 2-sinx=cos^2x +7sin^2x$

2. Originally Posted by sigma1
hello can someone please demonstrate how a question like this is solved for the the value of (x) between 0 and 360 degrees.

$\displaystyle 2-sinx=cos^2x +7sin^2x$
Start by substituting $\displaystyle \cos^2 x = 1 - \sin^2 x$. Then re-arrange the resulting equation into a quadratic equation where $\displaystyle \sin x$ is the variable. Solve this equation for $\displaystyle \sin x$ and then solve for x.

If you need more help, please show all your work and say where you get stuck.

3. ok i have

$\displaystyle 2-sinx =1-sin^2x =7sin^2x$
$\displaystyle 2-sinx= 1-6sin^2x$

$\displaystyle -sinx=-6sin^2x$

i cannot seem to get a value between 0 and 360 for my angle i suspect i did something wrong along the way... can you show me how its done..thanks..

4. Originally Posted by sigma1
ok i have

$\displaystyle 2-sinx =1-sin^2x =7sin^2x$
$\displaystyle 2-sinx= 1-6sin^2x$
$\displaystyle 2-sin(x)=1-6sin^{2}(x)\rightarrow-sin(x)=-1-6sin^{2}(x)$