1. ## Solving trig equations

i have a study guide for a quiz tomorrow, but i am not sure how to solve a few of the problems and feel pretty unprepared. it would be great if anyone could help me with these:

use the interval (0, 2 pi)
1. 2sin^2(theta)-sin(theta) = 1
2.(3cot^3)x-cotx= Square root of 3
3. (2sin^2)2theta = 1

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2. (12sin^2)x-5sinx-2 for the interval (0,360) in degrees

(im not sure exactly what the last two are asking for you to do, so hopefully someone recognizes it)
thank you

2. Originally Posted by Jimmie
1. 2sin^2(theta)-sin(theta) = 1
$2\sin^2\theta-\sin\theta = 1$

make $\sin\theta = a$

giving $2a^2-a=1 \implies 2a^2-a-1=0 \implies (2a+1)(a-1)=0$

now by the null factor law $a = \frac{-1}{2},1$

so yu now need to solve

$\sin\theta = \frac{-1}{2},1$

3. Originally Posted by pickslides
$2\sin^2\theta-\sin\theta = 1$

make $\sin\theta = a$

giving $2a^2-a=1 \implies 2a^2-a-1=0 \implies (2a+1)(a-1)=0$

now by the null factor law $a = \frac{-1}{2},1$

so yu now need to solve

$\sin\theta = \frac{-1}{2},1$
thank you, so all i need to do after that is inverse sin?

4. Originally Posted by Jimmie
thank you, so all i need to do after that is inverse sin?

You could do that, but that will not give you all values on $(0,2\pi)$ ?

I would use triangles to solve and identities from the unit circle for additional values.

5. Jimmie,
I have the first one for you:
2sin^2(theta) - sin(theta) - 1 = 0
(2sin(theta) + 1) (sin(theta) -1) = 0
using the zero factor rule, sin(theta) = 1 and -1/2
On the unit circle these sin values match with $\pi$/2, 7 $\pi$/6, and 11 $\pi$/6,
But because the angle was referred by the term "theta" I should give the answer in degrees. So the answers would be 90 degrees, 210 degrees and 330 degrees.
~sif