Step-by-step explanation:
We have,
[tex]\sin x=-\dfrac{\sqrt2}{2}\ ......(i)[/tex]
[tex]\cos x=\dfrac{\sqrt 2}{2}\ ......(ii)[/tex]
Applying [tex]\sin^{-1}[/tex] on both sides of equation (i) we get :
[tex]\sin^{-1} \sin x=\sin^{-1}(-\dfrac{\sqrt2}{2})[/tex]
We know that, [tex]\sin^{-1} \sin x=x[/tex]
So,
[tex]x=\sin^{-1}(-\dfrac{\sqrt2}{2})[/tex]
Also, [tex]\sin(-45)=\dfrac{-\sqrt2}{2}[/tex]
So,
[tex]x=-45^{\circ}[/tex]
Similarly,
Applying [tex]\cos^{-1}[/tex] on both sides of equation (ii) we get :
[tex]\cos^{-1} \cos x=\cos^{-1}(\dfrac{\sqrt2}{2})[/tex]
We know that, [tex]\cos^{-1} \cos x=x[/tex]
The cosine function is positive in the first and fourth quadrant.
So,
[tex]x=45^{\circ}[/tex]
