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...?
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 $\displaystyle \cos^2 \theta = 1 - \sin^2 \theta$ and re-arrange the resulting into a quadratic equation where $\displaystyle \sin \theta$ 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.
Substitute $\displaystyle w = \sin \theta$. Then, following from my earlier reply, you have:
$\displaystyle 5 - 7 w = 2(1 - w^2) \Rightarrow 2w^2 - 7w + 3 =0$.
Solve for w. One solution is rejected (why?). The other solution leads to $\displaystyle \sin \theta = \frac{1}{2}$. Solve this equation.
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.
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.