Trigonometric equations to solve for angles between 0° and 180°.

Here are the step-by-step solutions for each equation. We need to find all angles between $0^\circ$ and $180^\circ$.

a) $\sin 2x = 0.5$
1. Find the reference angle: $\sin^{-1}(0.5) = 30^\circ$.
2. Let $A = 2x$. Since $0^\circ \le x \le 180^\circ$, then $0^\circ \le A \le 360^\circ$.
3. In this range, sine is positive in Quadrants I and II.
* $A_1 = 30^\circ$
* $A_2 = 180^\circ - 30^\circ = 150^\circ$
4. Solve for $x$:
* $2x = 30^\circ \Rightarrow x = 15^\circ$
* $2x = 150^\circ \Rightarrow x = 75^\circ$

b) $\sin 3x = 1$
1. Find the reference angle: $\sin^{-1}(1) = 90^\circ$.
2. Let $A = 3x$. Range: $0^\circ \le A \le 540^\circ$.
3. Sine equals 1 only at $90^\circ$ within one cycle. We add $360^\circ$ to find other solutions in the extended range.
* $A_1 = 90^\circ$
* $A_2 = 90^\circ + 360^\circ = 450^\circ$
* ($90^\circ + 720^\circ = 810^\circ$ is too big)
4. Solve for $x$:
* $3x = 90^\circ \Rightarrow x = 30^\circ$
* $3x = 450^\circ \Rightarrow x = 150^\circ$

c) $\cos 2x = 0.5$
1. Reference angle: $\cos^{-1}(0.5) = 60^\circ$.
2. Let $A = 2x$. Range: $0^\circ \le A \le 360^\circ$.
3. Cosine is positive in Quadrants I and IV.
* $A_1 = 60^\circ$
* $A_2 = 360^\circ - 60^\circ = 300^\circ$
4. Solve for $x$:
* $2x = 60^\circ \Rightarrow x = 30^\circ$
* $2x = 300^\circ \Rightarrow x = 150^\circ$

d) $\cos(\frac{x}{2}) = 0.5$
1. Reference angle: $\cos^{-1}(0.5) = 60^\circ$.
2. Let $A = \frac{x}{2}$. Range: $0^\circ \le A \le 90^\circ$.
3. In this small range, there is only one solution where cosine is positive (Quadrant I).
* $A = 60^\circ$
4. Solve for $x$:
* $\frac{x}{2} = 60^\circ \Rightarrow x = 120^\circ$

e) $\tan 2x = 1$
1. Reference angle: $\tan^{-1}(1) = 45^\circ$.
2. Let $A = 2x$. Range: $0^\circ \le A \le 360^\circ$.
3. Tangent is positive in Quadrants I and III. The period is $180^\circ$.
* $A_1 = 45^\circ$
* $A_2 = 45^\circ + 180^\circ = 225^\circ$
4. Solve for $x$:
* $2x = 45^\circ \Rightarrow x = 22.5^\circ$
* $2x = 225^\circ \Rightarrow x = 112.5^\circ$ (Rounds to $113^\circ$)

f) $\cos 2\theta = -0.6$
1. Reference angle (ignore negative sign for now): $\cos^{-1}(0.6) \approx 53.13^\circ$.
2. Let $A = 2\theta$. Range: $0^\circ \le A \le 360^\circ$.
3. Cosine is negative in Quadrants II and III.
* $A_1 = 180^\circ - 53.13^\circ = 126.87^\circ$
* $A_2 = 180^\circ + 53.13^\circ = 233.13^\circ$
4. Solve for $\theta$:
* $2\theta = 126.87^\circ \Rightarrow \theta \approx 63.4^\circ$
* $2\theta = 233.13^\circ \Rightarrow \theta \approx 117^\circ$

g) $\tan 2\theta = -0.5$
1. Reference angle: $\tan^{-1}(0.5) \approx 26.57^\circ$.
2. Let $A = 2\theta$. Range: $0^\circ \le A \le 360^\circ$.
3. Tangent is negative in Quadrants II and IV.
* $A_1 = 180^\circ - 26.57^\circ = 153.43^\circ$
* $A_2 = 360^\circ - 26.57^\circ = 333.43^\circ$
4. Solve for $\theta$:
* $2\theta = 153.43^\circ \Rightarrow \theta \approx 76.7^\circ$
* $2\theta = 333.43^\circ \Rightarrow \theta \approx 167^\circ$

h) $\tan(\frac{\theta}{2}) = \sqrt{3}$
1. Reference angle: $\tan^{-1}(\sqrt{3}) = 60^\circ$.
2. Let $A = \frac{\theta}{2}$. Range: $0^\circ \le A \le 90^\circ$.
3. Tangent is positive in Quadrant I.
* $A = 60^\circ$
4. Solve for $\theta$:
* $\frac{\theta}{2} = 60^\circ \Rightarrow \theta = 120^\circ$

i) $\sin 4x = \frac{\sqrt{2}}{2}$
1. Reference angle: $\sin^{-1}(\frac{\sqrt{2}}{2}) = 45^\circ$.
2. Let $A = 4x$. Range: $0^\circ \le A \le 720^\circ$ (since $4 \times 180 = 720$).
3. Sine is positive in Quadrants I and II. We look for solutions every $360^\circ$.
* $A_1 = 45^\circ$
* $A_2 = 180^\circ - 45^\circ = 135^\circ$
* $A_3 = 45^\circ + 360^\circ = 405^\circ$
* $A_4 = 135^\circ + 360^\circ = 495^\circ$
* (Next ones would be over 720)
4. Solve for $x$:
* $4x = 45^\circ \Rightarrow x = 11.25^\circ$ (Rounds to $11.3^\circ$)
* $4x = 135^\circ \Rightarrow x = 33.75^\circ$ (Rounds to $33.8^\circ$)
* $4x = 405^\circ \Rightarrow x = 101.25^\circ$ (Rounds to $101^\circ$)
* $4x = 495^\circ \Rightarrow x = 123.75^\circ$ (Rounds to $124^\circ$)

j) $\tan(\frac{\theta}{3}) = 0.3$
1. Reference angle: $\tan^{-1}(0.3) \approx 16.70^\circ$.
2. Let $A = \frac{\theta}{3}$. Range: $0^\circ \le A \le 60^\circ$ (since $180/3 = 60$).
3. Tangent is positive in Quadrant I. The next positive solution would be $16.70 + 180$, which is outside our range for A.
* $A = 16.70^\circ$
4. Solve for $\theta$:
* $\frac{\theta}{3} = 16.70^\circ \Rightarrow \theta \approx 50.1^\circ$

Final Answer:
a) $15^\circ, 75^\circ$
b) $30^\circ, 150^\circ$
c) $30^\circ, 150^\circ$
d) $120^\circ$
e) $22.5^\circ, 113^\circ$
f) $63.4^\circ, 117^\circ$
g) $76.7^\circ, 167^\circ$
h) $120^\circ$
i) $11.3^\circ, 33.8^\circ, 101^\circ, 124^\circ$
j) $50.1^\circ$