# Thread: Precalculus Help. Finding exact solutions on the equation

1. ## Precalculus Help. Finding exact solutions on the equation

Hi

Find all exact solutions on [0,2pi)

2(cos^2)x + 3sinx = 3

Someone said i had to use (cos^2)x + (sin^2)x = 1 and use sin x as a variable, but i do not understand what that means.

Can someone explain to me step by step?

2. Exactly what that person said. Use sin x as a variable.

Replace $\displaystyle \cos^2 x$ with $\displaystyle 1 - \sin^2 x$:
\displaystyle \begin{aligned} 2\cos^2 x + 3\sin x &= 3 \\ 2(1 - \sin^2 x) + 3\sin x &= 3 \\ 2 - 2\sin^2 x + 3\sin x &= 3 \\ 2\sin^2 x - 3\sin x + 1 &= 0 \end{aligned}

See how this looks like a quadratic? If you replaced $\displaystyle \sin x$ with $\displaystyle y$ you would get
$\displaystyle 2y^2 - 3y + 1 = 0$

Solve the quadratic for y, plug back in $\displaystyle \sin x$ for y, and solve for x.

3. thank you!

so i factored out 2(sin^2)x - 3 sinx + 1 = 0

i got (sinx -1) (sin x - 1)

set them equal to zero and got sin x = 1

making x = pi/2

4. Originally Posted by danielh9103
so i factored out 2(sin^2)x - 3 sinx + 1 = 0

i got (sinx -1) (sin x - 1)
No, the factoring is not right (unless there is a typo)? If you multiply it out you get
$\displaystyle (\sin x - 1)(\sin x - 1) = \sin^2 x - 2\sin x + 1$

5. im sorry! it was a typo!

(2 sinx -1) (sinx - 1)

sinx= 1/2 sinx = 1

x = pi/2, pi/6, 5pi/6

6. Much better now!