Hello there!
Previously, we learnt that to solve the equation, we have to isolate the sin, cos, tan, etc first.
<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>
The first question has sin both sides. Notice that if we move sin(theta) to left. We get:-
![\displaystyle \large{2 {sin}^{2} \theta - sin \theta = 0}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B2%20%7Bsin%7D%5E%7B2%7D%20%20%5Ctheta%20-%20sin%20%5Ctheta%20%3D%200%7D)
We can common factor out the expression.
![\displaystyle \large{sin \theta(2sin \theta - 1) = 0}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7Bsin%20%5Ctheta%282sin%20%5Ctheta%20-%201%29%20%3D%200%7D)
It is a trigonometric equation in quadraric pattern.
We consider both equations:-
<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>E</u><u>q</u><u>u</u><u>a</u><u>t</u><u>i</u><u>o</u><u>n</u>
<u>
</u>
Remind that sin = y. When sin theta = 0. It means that it lies on the positive x-axis.
We know that 0 satisfies the equation, because sin(0) is 0.
Same goes for π as well, but 2π does not count because the interval is from 0 ≤ theta < 2π.
Hence:-
![\displaystyle \large { \theta = 0,\pi}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%20%7B%20%20%5Ctheta%20%3D%200%2C%5Cpi%7D)
<u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u> </u><u>E</u><u>q</u><u>u</u><u>a</u><u>t</u><u>i</u><u>o</u><u>n</u>
![\displaystyle \large{2sin \theta - 1 = 0}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Clarge%7B2sin%20%5Ctheta%20-%201%20%3D%200%7D)
First, as we learnt. We isolate sin.
![\displaystyle \large{sin \theta = \frac{1}{2} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7Bsin%20%5Ctheta%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%7D)
We know that, sin is positive in Quadrant 1 and 2.
As we learnt from previous question, we use π - (ref. angle) to find Q2 angle.
We know that sin(π/6) is 1/2. Hence π/6 is our reference angle. Since π/6 is in Q1, we only have to find Q2.
<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>u</u><u>a</u><u>d</u><u>r</u><u>a</u><u>n</u><u>t</u><u> </u><u>2</u>
![\displaystyle \large{\pi - \frac{\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} } \\ \displaystyle \large{ \frac{5\pi}{6} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B%5Cpi%20-%20%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20%20%3D%20%20%5Cfrac%7B6%5Cpi%7D%7B6%7D%20%20-%20%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20%7D%20%5C%5C%20%20%5Cdisplaystyle%20%5Clarge%7B%20%5Cfrac%7B5%5Cpi%7D%7B6%7D%20%7D)
Hence:-
![\displaystyle \large{ \theta = \frac{\pi}{6} , \frac{5\pi}{6} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B%20%5Ctheta%20%3D%20%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20%2C%20%5Cfrac%7B5%5Cpi%7D%7B6%7D%20%7D)
Since both first and second equations are apart of same equation. Therefore, mix both theta from first and second.
Therefore, the solutions to the first question:-
![\displaystyle \large \boxed{ \theta = 0,\pi, \frac{\pi}{6} , \frac{5\pi}{6} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%20%5Cboxed%7B%20%5Ctheta%20%3D%200%2C%5Cpi%2C%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20%2C%20%5Cfrac%7B5%5Cpi%7D%7B6%7D%20%7D)
<u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>
This one is a reciprocal of tan, also known as cot.
![\displaystyle \large{cot3 \theta = 1}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Clarge%7Bcot3%20%5Ctheta%20%3D%201%7D)
For this, I will turn cot to 1/tan.
![\displaystyle \large{ \frac{1}{tan3 \theta} = 1}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B%20%5Cfrac%7B1%7D%7Btan3%20%5Ctheta%7D%20%20%3D%201%7D)
Multiply whole equation by tan3 theta, to get rid of the denominator.
![\displaystyle \large{ \frac{1}{tan3 \theta} \times tan3 \theta = 1 \times tan3 \theta } \\ \displaystyle \large{ 1= tan3 \theta }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B%20%5Cfrac%7B1%7D%7Btan3%20%5Ctheta%7D%20%20%5Ctimes%20%20tan3%20%5Ctheta%20%3D%201%20%5Ctimes%20tan3%20%5Ctheta%20%7D%20%5C%5C%20%20%5Cdisplaystyle%20%5Clarge%7B%201%3D%20%20tan3%20%5Ctheta%20%7D)
We also learnt about how to deal with number beside theta.
We increase the interval, by multiplying with the number.
Since our interval is:-
![\displaystyle \large{0 \leqslant \theta < 2\pi}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B0%20%5Cleqslant%20%20%5Ctheta%20%3C%202%5Cpi%7D)
Multiply the whole interval by 3.
![\displaystyle \large{0 \times 3 \leqslant \theta \times 3 < 2\pi \times 3} \\ \displaystyle \large{0 \leqslant 3 \theta < 6\pi }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B0%20%5Ctimes%203%20%5Cleqslant%20%20%5Ctheta%20%5Ctimes%203%20%3C%202%5Cpi%20%5Ctimes%203%7D%20%5C%5C%20%20%5Cdisplaystyle%20%5Clarge%7B0%20%5Cleqslant%203%20%5Ctheta%20%20%3C%206%5Cpi%20%7D)
We also know that tan is positive in Quadrant 1 and Quadrant 3.
and tan(π/4) is 1. Therefore, π/4 is our reference angle and our first theta value.
When we want to find Quadrant 3, we use π + (ref. angle).
<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>
<u>
</u>
Hence, our theta values are π/4 and 5π/4. But that is for [0,2π) interval. We want to find theta values over [0,6π) interval.
As we learnt previously, that we use theta + 2πk to find values that are in interval greater than 2π.
As for tangent, we use:-
![\displaystyle \large{ \theta + \pi k = \theta}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B%20%5Ctheta%20%2B%20%5Cpi%20k%20%3D%20%20%5Ctheta%7D)
Because tan is basically a slope or line proportional graph. So it gives the same value every π period.
Now imagine a unit circle, and make sure to have some basic geometry knowledge. Know that when values addition by 180° or π would give a straight angle.
We aren't using k = 1 for this because we've already found Q3 angle.
Since we know Q1 and Q3 angle in [0,2π).
We can also use theta + 2πk if you want.
<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>V</u><u>a</u><u>l</u><u>u</u><u>e</u><u> </u><u>o</u><u>r</u><u> </u><u>π</u><u>/</u><u>4</u>
![\displaystyle \large{ \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} } \\ \displaystyle \large{ \frac{\pi}{4} + 4\pi = \frac{17\pi}{4} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%20%2B%202%5Cpi%20%3D%20%20%5Cfrac%7B9%5Cpi%7D%7B4%7D%20%7D%20%5C%5C%20%20%5Cdisplaystyle%20%5Clarge%7B%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%20%2B%204%5Cpi%20%3D%20%20%5Cfrac%7B17%5Cpi%7D%7B4%7D%20%7D)
<u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u> </u><u>V</u><u>a</u><u>l</u><u>u</u><u>e</u><u> </u><u>o</u><u>r</u><u> </u><u>5</u><u>π</u><u>/</u><u>4</u>
![\displaystyle \large{ \frac{5\pi}{4} + 2\pi = \frac{13\pi}{4} } \\ \displaystyle \large{ \frac{5\pi}{4} + 4\pi = \frac{21\pi}{4} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B%20%5Cfrac%7B5%5Cpi%7D%7B4%7D%20%20%2B%202%5Cpi%20%3D%20%20%5Cfrac%7B13%5Cpi%7D%7B4%7D%20%7D%20%5C%5C%20%20%5Cdisplaystyle%20%5Clarge%7B%20%5Cfrac%7B5%5Cpi%7D%7B4%7D%20%20%2B%204%5Cpi%20%3D%20%20%5Cfrac%7B21%5Cpi%7D%7B4%7D%20%7D)
Yes, I use theta + 2πk for finding other values.
Therefore:-
![\displaystyle \large{3 \theta = \frac{\pi}{4} , \frac{5\pi}{4} , \frac{9\pi}{4}, \frac{17\pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B3%20%5Ctheta%20%3D%20%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%20%2C%20%5Cfrac%7B5%5Cpi%7D%7B4%7D%20%2C%20%5Cfrac%7B9%5Cpi%7D%7B4%7D%2C%20%5Cfrac%7B17%5Cpi%7D%7B4%7D%20%2C%20%5Cfrac%7B13%5Cpi%7D%7B4%7D%20%2C%20%5Cfrac%7B21%5Cpi%7D%7B4%7D%20%20%7D)
Then we divide every values by 3.
![\displaystyle \large \boxed{\theta = \frac{\pi}{12} , \frac{5\pi}{12} , \frac{9\pi}{12}, \frac{17\pi}{12} , \frac{13\pi}{12} , \frac{21\pi}{12} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%20%5Cboxed%7B%5Ctheta%20%3D%20%20%5Cfrac%7B%5Cpi%7D%7B12%7D%20%20%2C%20%5Cfrac%7B5%5Cpi%7D%7B12%7D%20%2C%20%5Cfrac%7B9%5Cpi%7D%7B12%7D%2C%20%5Cfrac%7B17%5Cpi%7D%7B12%7D%20%2C%20%5Cfrac%7B13%5Cpi%7D%7B12%7D%20%2C%20%5Cfrac%7B21%5Cpi%7D%7B12%7D%20%20%7D)
Let me know if you have any questions!