4-6 Solve the equation for θ (0 ≤θ < 2π) . Be careful. There may be several
solutions. Write answers in radians.
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?
4-6 Solve the equation for θ (0 ≤θ < 2π) . Be careful. There may be several
solutions. Write answers in radians.
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?
Both 1 and 2 consists of the $\displaystyle \frac{\pi}{4}$ family...
For #1, you get two angles, $\displaystyle \vartheta=\frac{\pi}{4}$ 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:
$\displaystyle \tan^2\vartheta-\tan\vartheta=0\implies\tan\vartheta(\tan\vartheta-1)=0$
We will get two different equations:
$\displaystyle \tan\vartheta=0$ and $\displaystyle \tan\vartheta=1$
Keep in mind the restriction on the angle : $\displaystyle 0\leq\vartheta<2\pi$
Also keep in mind which quadrants tangent is positive!
I hope this helps.
--Chris
#3 : $\displaystyle \cos^2\vartheta+\sin\vartheta=1$
First convert the eqaution into sine. Use the fact that $\displaystyle \cos^2\vartheta=1-\sin^2\vartheta$
Thus, our equation then becomes $\displaystyle 1-\sin^2\vartheta+\sin\vartheta=1\implies \sin^2\vartheta-\sin\vartheta=0\implies \sin\vartheta(\sin\vartheta-1)=0$
Again, we get two equations:
$\displaystyle \sin\vartheta=0$ and $\displaystyle \sin\vartheta=1$
Keep in mind the restriction on the angle : $\displaystyle 0\leq\vartheta<2\pi$
I hope this helps.
--Chris
Erm...
I overlooked something pretty simple and thought of one equation, but there were two sets of equations:
$\displaystyle \sin\vartheta=\tfrac{1}{\sqrt{2}}$ and $\displaystyle \sin\vartheta=-\tfrac{1}{\sqrt{2}}$
So you should get four solutions, not two. Sorry about that! It was my mistake!
--Chris