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

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Solve the trigonometric equation for all values $$ 0 \leq x<2 \pi $$ $$ \sin ^{2} x-1=0 $$

UpStudy AI · Accepted Answer

Solve the equation $$ \sin^{2}(x)-1=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\sin^{2}\left(xight)-1=0$$
- step1: Move the constant to the right side:
  $$\sin^{2}\left(xight)=0+1$$
- step2: Remove 0:
  $$\sin^{2}\left(xight)=1$$
- step3: Simplify the expression:
  $$\sin\left(xight)=\pm \sqrt{1}$$
- step4: Simplify:
  $$\sin\left(xight)=\pm 1$$
- step5: Separate into possible cases:
  $$\begin{align}&\sin\left(xight)=1\&\sin\left(xight)=-1\end{align}$$
- step6: Calculate:
  $$\begin{align}&x=\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z}\&x=\frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step7: Find the union:
  $$x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
The solutions to the trigonometric equation $$ \sin^{2}x - 1 = 0 $$ for all values $$ 0 \leq x < 2\pi $$ are $$ x = \frac{\pi}{2} + k\pi $$, where $$ k $$ is an integer.
The solutions are $$ x = \frac{\pi}{2} $$ and $$ x = \frac{3\pi}{2} $$.

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

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