How do you solve the equation #sqrt2sinx-1=0# for #-360^circ<x<=360^circ#?

To solve the equation (\sqrt{2}\sin(x) - 1 = 0) for (-360^\circ \leq x \leq 360^\circ), follow these steps:

1. Add (1) to both sides of the equation: (\sqrt{2}\sin(x) = 1)

2. Divide both sides of the equation by (\sqrt{2}): (\sin(x) = \frac{1}{\sqrt{2}})

3. Find the angles in the specified range where the sine function equals (\frac{1}{\sqrt{2}}), which is (\frac{\pi}{4}) or (45^\circ) and (\frac{5\pi}{4}) or (225^\circ).

4. Since the range given is in degrees, ensure that the solutions are within the specified range, (-360^\circ \leq x \leq 360^\circ).

   Therefore, the solutions within the specified range are (x = 45^\circ) and (x = 225^\circ).

To solve the equation ( \sqrt{2}\sin(x) - 1 = 0 ) for ( -360^\circ < x \leq 360^\circ ), first, isolate the sine term. Then, solve for ( x ) by taking the inverse sine of both sides.

( \sqrt{2}\sin(x) - 1 = 0 )

Add 1 to both sides:

( \sqrt{2}\sin(x) = 1 )

Divide both sides by ( \sqrt{2} ):

( \sin(x) = \frac{1}{\sqrt{2}} )

Taking the inverse sine of both sides:

( x = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) )

( x = \frac{\pi}{4} ) or ( x = \frac{5\pi}{4} ) (in radians)

Since the sine function is positive in the first and third quadrants, and ( \frac{\pi}{4} ) and ( \frac{5\pi}{4} ) fall within the specified range ( -360^\circ < x \leq 360^\circ ), the solutions are:

( x = \frac{\pi}{4} ) or ( x = \frac{5\pi}{4} )