Question

What values for theta(0< or equal to theta < or equal to 2pi) satisfy the equation?
3sin theta = sin theta - 1

tan^2 theta = -3/2sec theta

Help! I have no idea what im doing

Answer 1

for the 1st equation

\[3\sin(\theta) = \sin(\theta) - 1\]

so start by subtracting \[\sin(\theta)\]

from both sides of the equation...

what do you get..?

Answer 2

ummm 4 sin=-1?

Answer 3

or 2sin=-1?

Answer 4

its the 2nd option

\[2 \sin(\theta) =-1\]

if you divide both sides of the equation by 2 you get

\[\sin(\theta) = -\frac{1}{2}\] does that make sense..?

Answer 5

yes...
so 2 would be
tan^2 = -3/2 sec
sin/cos ^2 = -3/2 1/cos = sin^2/cos=-3/2 ?

Answer 6

well getting back to the 1st equation...

you can now solve this... find

sin = 1/2 and sin is negative in the 3rd and 4th quadrants so

the answer is 180 + theta and 360 - theta

hope that is ok..

Answer 7

pi/3 and 5pi/3

Answer 8

??

Answer 9

for the 2nd equation rewrite it as

\[\frac{\sin^2(\theta)}{\cos^2(\theta)} = -\frac{3}{2\cos(\theta)}\]

multiply both sides by \[\cos^2(\theta)\]

and you get

\[\sin^2(\theta) = \frac{-3\cos(\theta)}{2}\]

then using sin^2 + cos^2 = 1 you can make the substitution

\[1 - \cos^2(\theta) = -\frac{3\cos(\theta)}{2} \]

which becomes

\[\cos^2(\theta) - \frac{3\cos(\theta)}{2} - 1 = 0\]

so you can solve the quadratic

hope it makes sense

Answer 10

ok... so if you are working in radians

the angles are

\[\pi + \theta~~and~~~2\pi - \theta\]

Answer 11

Now you lost me completely.

Answer 12

remember from an exact value triangle

\[\sin(\theta) = \frac{1}{2}~~~then~~~~\theta = \frac{\pi}{6}\]

Answer 13

But there must also be a second one... so pi/6 and 3pi/6?

Answer 14

here is the diagram



since sin is negative it means the angles are in the 3rd and 4th quadrants...

so the 3rd quadrant angle is

\[\pi + \frac{\pi}{6}\]

and the 4th quadrant angle is
\[2\pi - \theta\]