Solve the trigonometric equation for all values $$ 0 \leq x

Question

Solve the trigonometric equation for all values $$ 0 \leq x<2 \pi $$ $$ \sin x+1=0 $$

UpStudy AI · Accepted Answer

**Step 1.** Write the equation in its simplified form.  
$$
\sin x + 1 = 0 \quad \Rightarrow \quad \sin x = -1.
$$

**Step 2.** Identify the general solution for $$\sin x = -1$$.  
The sine function equals $$-1$$ at:
$$
x = \frac{3\pi}{2} + 2\pi k, \quad \text{where } k \in \mathbb{Z}.
$$

**Step 3.** Find the solutions within $$0 \leq x < 2\pi$$.  
Substitute $$k = 0$$ into the general solution:
$$
x = \frac{3\pi}{2}.
$$

For any other integer $$k$$, the value would be outside the interval $$0 \leq x < 2\pi$$:
- If $$k = 1$$, then $$x = \frac{3\pi}{2} + 2\pi = \frac{7\pi}{2}$$, which is greater than $$2\pi$$.
- If $$k = -1$$, then $$x = \frac{3\pi}{2} - 2\pi = -\frac{\pi}{2}$$, which is less than $$0$$.

Thus, the only solution in the interval is:
$$
x = \frac{3\pi}{2}.
$$

**Final Answer.**  
$$
\boxed{x = \frac{3\pi}{2}}
$$
The solution to the equation $$\sin x + 1 = 0$$ within $$0 \leq x < 2\pi$$ is: $$ x = \frac{3\pi}{2} $$

Solve the trigonometric equation for all values \( 0 \leq x<2 \pi \) \[ \sin x+1=0 \]

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