# Math Help - trigonometric equation

1. ## trigonometric equation

4-6 Solve the equation for θ (0 ≤θ < 2π) . Be careful. There may be several
4. 2sin^2(θ) =1 5. tan^2(θ) − tan(θ) = 0 7. cos^2(θ) + sin(θ) =1

is the first and second ones just in the 45 degree family?

2. Originally Posted by CalcGeek31
4-6 Solve the equation for θ (0 ≤θ < 2π) . Be careful. There may be several
4. 2sin^2(θ) =1 5. tan^2(θ) − tan(θ) = 0 7. cos^2(θ) + sin(θ)

is the first and second ones just in the 45 degree family?
Both 1 and 2 consists of the $\frac{\pi}{4}$ family...

For #1, you get two angles, $\vartheta=\frac{\pi}{4}$ and another angle. Just keep in mind the quadrants where sine is positive! I'll leave it for you to find the other angle.

However, for #2, we see that we can manipulate the equation a little bit:

$\tan^2\vartheta-\tan\vartheta=0\implies\tan\vartheta(\tan\vartheta-1)=0$

We will get two different equations:

$\tan\vartheta=0$ and $\tan\vartheta=1$

Keep in mind the restriction on the angle : $0\leq\vartheta<2\pi$

Also keep in mind which quadrants tangent is positive!

I hope this helps.

--Chris

3. what about the third one? it is = 1 will go edit that

4. Originally Posted by CalcGeek31
4-6 Solve the equation for θ (0 ≤θ < 2π) . Be careful. There may be several
4. 2sin^2(θ) =1 5. tan^2(θ) − tan(θ) = 0 7. cos^2(θ) + sin(θ) =1

is the first and second ones just in the 45 degree family?
#3 : $\cos^2\vartheta+\sin\vartheta=1$

First convert the eqaution into sine. Use the fact that $\cos^2\vartheta=1-\sin^2\vartheta$

Thus, our equation then becomes $1-\sin^2\vartheta+\sin\vartheta=1\implies \sin^2\vartheta-\sin\vartheta=0\implies \sin\vartheta(\sin\vartheta-1)=0$

Again, we get two equations:

$\sin\vartheta=0$ and $\sin\vartheta=1$

Keep in mind the restriction on the angle : $0\leq\vartheta<2\pi$

I hope this helps.

--Chris

5. for #1... I just re did it 4 times and got 4 answers not 2 can anyone confirm?

6. Originally Posted by CalcGeek31
for #1... I just re did it 4 times and got 4 answers not 2 can anyone confirm?
Erm...

I overlooked something pretty simple and thought of one equation, but there were two sets of equations:

$\sin\vartheta=\tfrac{1}{\sqrt{2}}$ and $\sin\vartheta=-\tfrac{1}{\sqrt{2}}$

So you should get four solutions, not two. Sorry about that! It was my mistake!

--Chris