Option C:
x = 90°
Solution:
Given equation:
[tex]\sin x=1+\cos ^{2} x[/tex]
To find the degree:
[tex]\sin x=1+\cos ^{2} x[/tex]
Subtract 1 + cos²x from both sides.
[tex]\sin x-1-\cos ^{2} x=0[/tex]
Using the trigonometric identity:[tex]\cos ^{2}(x)=1-\sin ^{2}(x)[/tex]
[tex]\sin x-1-\left(1-\sin ^{2}x\right)=0[/tex]
[tex]\sin x-1-1+\sin ^{2}x=0[/tex]
[tex]\sin x-2+\sin ^{2}x=0[/tex]
[tex]\sin ^{2}x+\sin x-2=0[/tex]
Let sin x = u
[tex]u^2+u-2=0[/tex]
Factor the quadratic equation.
[tex](u+2)(u-1)=0[/tex]
u + 2 = 0, u – 1 = 0
u = –2, u = 1
That is sin x = –2, sin x = 1
sin x can't be smaller than –1 for real solutions. So ignore sin x = –2.
sin x = 1
The value of sin is 1 for 90°.
x = 90°.
Option C is the correct answer.
