Question

Solve the equation for principal values of x. Express solutions in degrees.

Answer 1

Answer:

C

Step-by-step explanation:

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Answer 2

Option C:

x = 90°

Solution:

Given equation:

$\sin x=1+\cos ^{2} x$

To find the degree:

$\sin x=1+\cos ^{2} x$

Subtract 1 + cos²x from both sides.

$\sin x-1-\cos ^{2} x=0$

Using the trigonometric identity:$\cos ^{2}(x)=1-\sin ^{2}(x)$

$\sin x-1-\left(1-\sin ^{2}x\right)=0$

$\sin x-1-1+\sin ^{2}x=0$

$\sin x-2+\sin ^{2}x=0$

$\sin ^{2}x+\sin x-2=0$

Let sin x = u

$u^2+u-2=0$

Factor the quadratic equation.

$(u+2)(u-1)=0$

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.