Solve the following equations given that 0<θ<360
a) (3cot^2)θ+1=5csc30
b) tan^2θ=1/3
$\displaystyle 3 \cot^2 \theta + 1 = 5 \csc 30$
$\displaystyle \Rightarrow \cot \theta = \pm \sqrt{\frac{5 \csc 30 - 1}3}$
$\displaystyle \Rightarrow \cot \theta = \pm \sqrt{\frac{\frac 52 - 1}3}$
now what?
$\displaystyle \tan^2 \theta = \frac 13$b) tan^2θ=1/3
$\displaystyle \Rightarrow \tan \theta = \pm \frac 1{\sqrt{3}}$
now what?
a) (3cot^2)θ+1=5csc30
That's the one posted. Correct and re-write it.
As posted, it defies any ammount of head-scratching.
b) tan^2θ=1/3
If that is
tan^2(theta) = 1/3, then,
Take the square root of both sides,
tan(theta) = +,- 1 / sqrt(3)
Plus or minus tan value. Meaning, in every quadrant, there should be theta.
tan(theta) = 1 / sqrt(3) is for theta = 30deg. That's in the 1st quadrant.
For the 2nd quadrant, theta = 180 -30 = 150 deg
For the 3rd quadrant, theta = 180 +30 = 210 deg
For the 4th quadrant, theta = 360 -30 = 330 deg
So,
theta = 30, 150, 210, 330 degrees.