MATH SOLVE

1 month ago

Q:
# Find a numerical value of one trigonometric function of x for cos^2x+ 2sin x-2=0

Accepted Solution

A:

Answer:x = 90°Step-by-step explanation:We are given a trigonometric function of x from which we have to a solution for x.
The function is [tex]\cos^{2} x + 2\sin x - 2 = 0[/tex]
⇒ [tex]1 - \sin^{2} x + 2\sin x - 2 = 0[/tex] {Since we know the identity [tex]\sin^{2} \alpha + \cos^{2} \alpha = 1[/tex]}
⇒ [tex]\sin^{2} x - 2 \sin x + 1 = 0[/tex]
⇒ [tex](\sin x - 1)^{2} = 0[/tex]
{Since we know the formula (a - b)² = a² - 2ab + b²}
⇒ [tex](\sin x - 1) = 0[/tex]
⇒ [tex]\sin x = 1 = \sin 90[/tex]
⇒ x = 90° (Answer)