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alexgriva [62]
3 years ago
11

Help me please thank you

Mathematics
1 answer:
Snezhnost [94]3 years ago
3 0
Point A is at (-5.1 , 3.3)
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2088291

_______________


Solve the trigonometric equation:

\mathsf{sin^2\,x-cos^2\,x=0}\qquad\qquad\mathsf{x\in [0,\,2\pi).}\\\\ \mathsf{(1-cos^2\,x)-cos^2\,x=0}\\\\ \mathsf{1-2\,cos^2\,x=0}\\\\ \mathsf{1=2\,cos^2\,x}\\\\ \mathsf{cos^2\,x=\dfrac{1}{2}}

\mathsf{cos\,x=\pm\,\sqrt{\dfrac{1}{2}}}\\\\\\ \mathsf{cos\,x=\pm\,\dfrac{1}{\sqrt{2}}}\\\\\\ \mathsf{cos\,x=\pm\,\dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}}\\\\\\ \mathsf{cos\,x=\pm\,\dfrac{\sqrt{2}}{2}}


\begin{array}{lcl} \footnotesize\begin{array}{l}\bullet\end{array}~~\mathsf{cos\,x=-\,\dfrac{\sqrt{2}}{2}}&\quad\Rightarrow\quad&\mathsf{x=\pi-\dfrac{\pi}{4}~~or~~x=\pi+\dfrac{\pi}{4}}\\\\ && \mathsf{x=\dfrac{4\pi}{4}-\dfrac{\pi}{4}~~or~~x=\dfrac{4\pi}{4}+\dfrac{\pi}{4}}\\\\ && \mathsf{x=\dfrac{3\pi}{4}~~or~~x=\dfrac{5\pi}{4}}\qquad\quad\checkmark\\\\\\ \footnotesize\begin{array}{l}\bullet\end{array}~~\mathsf{cos\,x=\,\dfrac{\sqrt{2}}{2}}&\quad\Rightarrow\quad&\mathsf{x=\dfrac{\pi}{4}~~or~~x=2\pi-\dfrac{\pi}{4}}\\\\ && \mathsf{x=\dfrac{\pi}{4}~~or~~x=\dfrac{8\pi}{4}-\dfrac{\pi}{4}}\\\\ && \mathsf{x=\dfrac{\pi}{4}~~or~~x=\dfrac{7\pi}{4}}\qquad\quad\checkmark \end{array}


Solution set:  \mathsf{S=\left\{\dfrac{\pi}{4},\,\dfrac{3\pi}{4},\,\dfrac{5\pi}{4},\,\dfrac{7\pi}{4} \right\}.}


I hope this helps. =)


Tags:   <em>solve trigonometric trig equation sine cosine sin cos identity trigonometry</em>

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