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sergiy2304 [10]
3 years ago
15

General solutions of sin(x-90)+cos(x+270)=-1 {both 90 and 270 are in degrees}

Mathematics
1 answer:
mixer [17]3 years ago
4 0

Answer:

\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.

Step-by-step explanation:

Given:

\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1

First, note that

\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x

So, the equation is

-\cos x+\sin x= -1

Multiply this equation by \frac{\sqrt{2}}{2}:

-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}

The general solution is

x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.

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