Example 2.4.2. Given $$ \sin 2 heta=\frac{-1}{\sqrt{2}} $$, solve for $$ heta $$. Example 2.4.3. Given $$ 2 \sin ^{2} heta+\sin heta-1=0 $$, solve for $$ heta $$.

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Example 2.4.2. Given $$ \sin 2 	heta=\frac{-1}{\sqrt{2}} $$, solve for $$ 	heta $$. Example 2.4.3. Given $$ 2 \sin ^{2} 	heta+\sin 	heta-1=0 $$, solve for $$ 	heta $$.

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**Example 2.4.2**

We start with the equation
$$
\sin 2\theta = \frac{-1}{\sqrt{2}}.
$$
Let
$$
\alpha = 2\theta.
$$
Then the equation becomes
$$
\sin \alpha = \frac{-1}{\sqrt{2}}.
$$

The sine function equals $$\frac{-1}{\sqrt{2}}$$ when
$$
\alpha = \frac{5\pi}{4} + 2\pi k \quad \text{or} \quad \alpha = \frac{7\pi}{4} + 2\pi k,\quad k\in\mathbb{Z}.
$$

Substitute back $$\alpha = 2\theta$$:
$$
2\theta = \frac{5\pi}{4} + 2\pi k \quad \text{or} \quad 2\theta = \frac{7\pi}{4} + 2\pi k.
$$

Dividing by 2 in each case gives:
$$
\theta = \frac{5\pi}{8} + \pi k \quad \text{or} \quad \theta = \frac{7\pi}{8} + \pi k,\quad k\in\mathbb{Z}.
$$

---

**Example 2.4.3**

We start with the equation
$$
2 \sin^2 \theta + \sin \theta - 1 = 0.
$$
Let
$$
u = \sin \theta.
$$
Then the equation becomes a quadratic in $$u$$:
$$
2u^2 + u - 1 = 0.
$$

To factor the quadratic, we look for two numbers that multiply to $$(2)(-1) = -2$$ and add to $$1$$. Notice that
$$
2u^2 + u - 1 = (2u - 1)(u + 1) = 0.
$$

Set each factor to zero:

1. $$2u - 1 = 0$$ leads to
   $$
   u = \frac{1}{2}.
   $$
   Therefore,
   $$
   \sin \theta = \frac{1}{2}.
   $$
   The general solutions where the sine is $$\frac{1}{2}$$ are:
   $$
   \theta = \frac{\pi}{6} + 2\pi k \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2\pi k,\quad k\in\mathbb{Z}.
   $$

2. $$u + 1 = 0$$ leads to
   $$
   u = -1.
   $$
   Therefore,
   $$
   \sin \theta = -1.
   $$
   The sine function equals $$-1$$ at:
   $$
   \theta = \frac{3\pi}{2} + 2\pi k,\quad k\in\mathbb{Z}.
   $$

Thus, the complete solution for Example 2.4.3 is:
$$
\theta = \frac{\pi}{6} + 2\pi k,\quad \theta = \frac{5\pi}{6} + 2\pi k,\quad \theta = \frac{3\pi}{2} + 2\pi k,\quad k\in\mathbb{Z}.
$$
**Example 2.4.2** Solutions for $$ \sin 2\theta = \frac{-1}{\sqrt{2}} $$ are: $$ \theta = \frac{5\pi}{8} + \pi k \quad \text{or} \quad \theta = \frac{7\pi}{8} + \pi k,\quad k\in\mathbb{Z}. $$ **Example 2.4.3** Solutions for $$ 2 \sin^2 \theta + \sin \theta - 1 = 0 $$ are: $$ \theta = \frac{\pi}{6} + 2\pi k,\quad \theta = \frac{5\pi}{6} + 2\pi k,\quad \theta = \frac{3\pi}{2} + 2\pi k,\quad k\in\mathbb{Z}. $$

Example 2.4.2. Given \( \sin 2 \theta=\frac{-1}{\sqrt{2}} \), solve for \( \theta \). Example 2.4.3. Given \( 2 \sin ^{2} \theta+\sin \theta-1=0 \), solve for \( \theta \).

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