Find the general solutions of the following equations. $$ \begin{array}{ll} ext { (a) } \sqrt{3} \cos heta-3 \sin heta=0 & ext { (b) } 2 \cos ^{2} heta=\cos heta \ ext { (c) }(\cos heta+\sin heta)(1-2 \sin heta)=0 & ext { (d) } 2 \sin ^{2} heta-\sin heta=1\end{array} $$

Question

Find the general solutions of the following equations. $$ \begin{array}{ll}	ext { (a) } \sqrt{3} \cos 	heta-3 \sin 	heta=0 & 	ext { (b) } 2 \cos ^{2} 	heta=\cos 	heta \ 	ext { (c) }(\cos 	heta+\sin 	heta)(1-2 \sin 	heta)=0 & 	ext { (d) } 2 \sin ^{2} 	heta-\sin 	heta=1\end{array} $$

UpStudy AI · Accepted Answer

(a) Start with
$$
\sqrt{3} \cos \theta - 3 \sin \theta = 0.
$$
Assume $$\cos \theta \neq 0$$ and divide through by $$\cos \theta$$:
$$
\sqrt{3} - 3 \tan \theta = 0 \quad \Longrightarrow \quad \tan \theta = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}.
$$
Since
$$
\tan \theta = \frac{1}{\sqrt{3}}
$$
has a principal value $$\theta = \pi/6$$, the general solution is
$$
\theta = \frac{\pi}{6} + k\pi,\quad k\in \mathbb{Z}.
$$

---

(b) The equation is
$$
2 \cos^2 \theta = \cos \theta.
$$
Bring all terms to one side:
$$
2 \cos^2 \theta - \cos \theta = 0.
$$
Factor out $$\cos \theta$$:
$$
\cos \theta (2\cos \theta - 1) = 0.
$$
This gives two cases:

1. $$\cos \theta = 0$$. The general solution is
   $$
   \theta = \frac{\pi}{2} + k\pi,\quad k\in \mathbb{Z}.
   $$
2. $$2\cos \theta - 1 = 0$$ which implies
   $$
   \cos \theta = \frac{1}{2}.
   $$
   The solutions for $$\cos \theta = \frac{1}{2}$$ are
   $$
   \theta = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad \theta = -\frac{\pi}{3} + 2k\pi,\quad k\in \mathbb{Z}.
   $$

---

(c) The equation is
$$
(\cos \theta + \sin \theta)(1 - 2 \sin \theta) = 0.
$$
So either
$$
\cos \theta + \sin \theta = 0
$$
or
$$
1 - 2\sin \theta = 0.
$$

For $$\cos \theta + \sin \theta = 0$$:
$$
\sin \theta = -\cos \theta.
$$
Dividing by $$\cos \theta$$ (assuming $$\cos \theta \neq 0$$):
$$
\tan \theta = -1.
$$
Thus,
$$
\theta = -\frac{\pi}{4} + k\pi,\quad k\in \mathbb{Z}.
$$

For $$1 - 2\sin \theta = 0$$:
$$
\sin \theta = \frac{1}{2}.
$$
This gives the general solutions:
$$
\theta = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2k\pi,\quad k\in \mathbb{Z}.
$$

---

(d) The equation is
$$
2\sin^2 \theta - \sin \theta = 1.
$$
Rearrange to obtain a quadratic equation in $$\sin \theta$$:
$$
2\sin^2 \theta - \sin \theta - 1 = 0.
$$
Let $$x = \sin \theta$$. Then
$$
2x^2 - x - 1 = 0.
$$
Factor the quadratic:
$$
(2x + 1)(x - 1) = 0.
$$
Thus, either
$$
2x + 1 = 0 \quad \text{or} \quad x - 1 = 0.
$$

1. $$2x + 1 = 0$$ gives
   $$
   \sin \theta = -\frac{1}{2}.
   $$
   The solutions for $$\sin \theta = -\frac{1}{2}$$ are
   $$
   \theta = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad \theta = \frac{11\pi}{6} + 2k\pi,\quad k\in \mathbb{Z}.
   $$
2. $$x - 1 = 0$$ gives
   $$
   \sin \theta = 1,
   $$
   so
   $$
   \theta = \frac{\pi}{2} + 2k\pi,\quad k\in \mathbb{Z}.
   $$

---

Final answers:

(a) $$\theta = \frac{\pi}{6} + k\pi,\quad k\in \mathbb{Z}$$.

(b) $$\theta = \frac{\pi}{2} + k\pi,\quad k\in \mathbb{Z}$$ or $$\theta = \frac{\pi}{3} + 2k\pi$$ or $$\theta = -\frac{\pi}{3} + 2k\pi,\quad k\in \mathbb{Z}$$.

(c) $$\theta = -\frac{\pi}{4} + k\pi,\quad k\in \mathbb{Z}$$ or $$\theta = \frac{\pi}{6} + 2k\pi$$ or $$\theta = \frac{5\pi}{6} + 2k\pi,\quad k\in \mathbb{Z}$$.

(d) $$\theta = \frac{7\pi}{6} + 2k\pi$$ or $$\theta = \frac{11\pi}{6} + 2k\pi$$ or $$\theta = \frac{\pi}{2} + 2k\pi,\quad k\in \mathbb{Z}$$.
(a) $$\theta = \frac{\pi}{6} + k\pi$$, where $$k$$ is any integer. (b) $$\theta = \frac{\pi}{2} + k\pi$$ or $$\theta = \frac{\pi}{3} + 2k\pi$$ or $$\theta = -\frac{\pi}{3} + 2k\pi$$, where $$k$$ is any integer. (c) $$\theta = -\frac{\pi}{4} + k\pi$$ or $$\theta = \frac{\pi}{6} + 2k\pi$$ or $$\theta = \frac{5\pi}{6} + 2k\pi$$, where $$k$$ is any integer. (d) $$\theta = \frac{7\pi}{6} + 2k\pi$$ or $$\theta = \frac{11\pi}{6} + 2k\pi$$ or $$\theta = \frac{\pi}{2} + 2k\pi$$, where $$k$$ is any integer.

Find the general solutions of the following equations. \( \begin{array}{ll}\text { (a) } \sqrt{3} \cos \theta-3 \sin \theta=0 & \text { (b) } 2 \cos ^{2} \theta=\cos \theta \\ \text { (c) }(\cos \theta+\sin \theta)(1-2 \sin \theta)=0 & \text { (d) } 2 \sin ^{2} \theta-\sin \theta=1\end{array} \)

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