Mathematics: Sin^2x-cos^2x=0 solve for the equation for the interval [0, 2pi)

Sin^2x-cos^2x=0 solve for the equation for the

Ответ:

lokijuhygtfrdfghj

06.01.2022

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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: solve trigonometric trig equation sine cosine sin cos identity trigonometry

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Ответ:

willijav000

06.03.2021

Mathematics

Step-by-step explanation:

Perimeter= 2( length+ width)

l= w+ 3

perimeter= 2(w+3+w)

30= 4w +6

4w= 24

w= 6

width = 6m

length = 6+3 = 9 m

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Sin^2x-cos^2x=0
solve for the equation for the interval [0, 2pi)

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