Precalculus Help. Finding exact solutions on the equation

Jul 2010
20
1
New Jersey
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?
 
Jan 2010
278
138
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.
 
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Jul 2010
20
1
New Jersey
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
 
Jan 2010
278
138
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
\(\displaystyle (\sin x - 1)(\sin x - 1) = \sin^2 x - 2\sin x + 1\)
:confused:
 
Jul 2010
20
1
New Jersey
im sorry! it was a typo!

(2 sinx -1) (sinx - 1)

sinx= 1/2 sinx = 1

x = pi/2, pi/6, 5pi/6
 
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