Question

Find, to the nearest tenth of a degree, all values of x for the interval 0 is less then or equal to x which is less then or equal to 360 that satisfy the equation 5sin^2x-9cosx-3=0

(REFER TO PICTURE ATTACHED)

Answer 1

x = 78.5° or x = 281.5°

Explanation:

5sin^2x-9cosx-3=0 ⇔ 5(1 -cos^2x) - 9cosx - 3 = 0

⇔ -5cos^2x - 9cosx + 2 = 0

⇔ 5cos^2x + 9cosx - 2 = 0

now we need to solve : 5x² - 9x + 2 = 0

5x² - 9x + 2 = 0 ⇔ 5(x - 0.2)(x + 2) = 0

⇔ x = 0.2 or x = -2

cosx = -2 (impossible)

cosx = 0.2 ⇔ cos⁻¹(0.2) = 78.463040967184

cos(360-78.463040967184) = 0.2

360-78.463040967184 = 281.536959032816