Help w/ this trig equation

[kpx001]

how would i go about solving 2sin^2 - 3sin + 1 = 0
i tried sin (2sin - 3 ) + 1
sin (2sin -3 ) = -1
then solved from there.
i need to get pi/6, pi/2, and 5pi/6 some how.

 

[fred2028]

how would i go about solving 2sin^2 - 3sin + 1 = 0
i tried sin (2sin - 3 ) + 1
sin (2sin -3 ) = -1
then solved from there.
i need to get pi/6, pi/2, and 5pi/6 some how.

This is a quadratic trigonometrical equation. How can you get 3 answers?
Anyways factor the equation using the quadratic formula

sin(x) = [-b +/- sqrt(b^2 - 4ac)] / 2a

That should give u 2 numbers. Or roots. So 2 x-values.
Then do the inverse functions.

 

[galactus]

That's because \(\displaystyle sin(\frac{5{\pi}}{6})=1/2\) and \(\displaystyle sin(\frac{\pi}{6})=1/2\)

 

[Mrspi]

how would i go about solving 2sin^2 - 3sin + 1 = 0
i tried sin (2sin - 3 ) + 1
sin (2sin -3 ) = -1
then solved from there.
i need to get pi/6, pi/2, and 5pi/6 some how.

I have no idea what you mean when you say "then solved from there."

Factor the left side:

(2 sin x - 1)(sin x - 1) = 0

Set each factor equal to 0 and solve.

If 2 sin x - 1 = 0, then
2 sin x = 1
sin x = 1/2
In the interval 0 < x < 2pi, when sin x = 1/2, then x = pi / 6 or x = (5 pi) / 6

If sin x - 1 = 0, then
sin x = 1
In the interval 0 < x <, when sin x = 1, then x = pi / 2

I hope this helps you.