Question

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

Answer 1

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

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
`(√(2))/(2):`

`-(√(2))/(2)\cos x+(√(2))/(2)\sin x= -(√(2))/(2)\\ \\(√(2))/(2)\cos x-(√(2))/(2)\sin x=(√(2))/(2)\\ \\\cos 45^(\circ)\cos x-\sin 45^(\circ)\sin x=(√(2))/(2)\\ \\\cos (x+45^(\circ))=(√(2))/(2)`

The general solution is

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