# Solving trig equations

• Mar 21st 2010, 01:40 PM
Jimmie
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 (Happy)
• Mar 21st 2010, 01:54 PM
pickslides
Quote:

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

$\sin\theta = \frac{-1}{2},1$
• Mar 21st 2010, 02:04 PM
Jimmie
Quote:

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?
• Mar 21st 2010, 02:08 PM
pickslides
Quote:

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.
• Mar 21st 2010, 02:57 PM
sifaka
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