1. ## Funny double-equation problem

5-7sin θ€ =2cos^2 θ€ , θ€ [-360°, 450°]

I'm just a little lost with the multiple use of Theda and €...

Just thought of something... is θ€ simply another way of writing θ? The € sign has no input on what the problem means...?

2. Originally Posted by wiseguy
5-7sin θ€ =2cos^2 θ€ , θ€ [-360°, 450°]

I'm just a little lost with the multiple use of Theda and €...

Just thought of something... is θ€ simply another way of writing θ? The € sign has no input on what the problem means...?
Solutions are required over the interval [-360 degrees, 450 degrees].

Substitute $\cos^2 \theta = 1 - \sin^2 \theta$ and re-arrange the resulting into a quadratic equation where $\sin \theta$ is the unknown.

If you need more help, please show all your work and say where you get stuck.

Also, please don't put questions in quote tags - it makes it too difficult to quote the question when replying.

3. Here's what I got...

5=2cos^2θ/7sinθ
(7/2)5=cos^2θ/sinθ
17.5=cosθ*cotθ

Can I carry this out like a normal equation? What I use to eliminate the cosx cotx mess?

4. I did an alternative approach to the problem, however I'm not sure if the 1 on the right side works

5=2cos^2θ/7sinθ
(7/2)5=1-sin^2θ/sinθ
17.5=1-sinθ

5. Originally Posted by wiseguy
I did an alternative approach to the problem, however I'm not sure if the 1 on the right side works

5=2cos^2θ/7sinθ
(7/2)5=1-sin^2θ/sinθ
17.5=1-sinθ
Substitute $w = \sin \theta$. Then, following from my earlier reply, you have:

$5 - 7 w = 2(1 - w^2) \Rightarrow 2w^2 - 7w + 3 =0$.

Solve for w. One solution is rejected (why?). The other solution leads to $\sin \theta = \frac{1}{2}$. Solve this equation.

6. Okay, I think I got it:

x=1/2, x=3

sinθ=1/2, sinθ=3
arcsin(1/2)=0.523599, arcsin3=no solution

so there is only one solution, and it is θ=0.523599 ...?

Thank you

7. Originally Posted by wiseguy
Okay, I think I got it:

x=1/2, x=3

sinθ=1/2, sinθ=3
arcsin(1/2)=0.523599, arcsin3=no solution

so there is only one solution, and it is θ=0.523599 ...?

Thank you
1) θ needs to be in degrees, as stated in the original problem, so θ = 30°.

2) In the future, you should express radian measures as something times pi if you can. IOW it's better to say θ = π/6 instead of 0.523599.... sinθ = 1/2 -> θ = π/6 or 30° is one of those things you should really memorize.

8. Originally Posted by wiseguy
Okay, I think I got it:

x=1/2, x=3

sinθ=1/2, sinθ=3
arcsin(1/2)=0.523599, arcsin3=no solution

so there is only one solution, and it is θ=0.523599 ...?

Thank you
Your answer is supposed to be in degrees since the question uses degree.

arcsin (0.5) = 30 degree and this is one of the solutions in the range given.

There are more: 150, 390, -330 , -210

9. Okay, how would I tie the thing where sin is positive in the first and second quadrant to the four solutions of 150, 390, -330 , -210?

10. Originally Posted by wiseguy
Okay, how would I tie the thing where sin is positive in the first and second quadrant to the four solutions of 150, 390, -330 , -210?
The reference angle is 30(1st quadrant) , 150(2nd quadrant). Note also that the period of a sin graph is 360. In other words, it repeats itself every 360.

so 30+360=390

How about 150+360=510? Look at the range.

Now you go clockwise direction, where sin is now positive in the 3rd and 4th quadrant.

In the 3rd quadrant, -(180+30) and the 4th: -(360-30)

As an alternative, you can use the general formula for sine.

11. Got it!

Now I have to remember this stuff... lol