Question

Can somebody please help me on this question?

14. Solve the following equations. Give your answers in the
interval 0º < x < 360°.
d) 1 + 3 sin x=0

Answer 1

Answer:

x ≈ 199.5°, x ≈ 340.5°

Step-by-step explanation:

Given

1 + 3sinx = 0 ( subtract 1 from both sides )

3sinx = - 1 ( divide both sides by 3 )

sinx = - $\frac{1}{3}$

Since sinx < 0 then x is in the third/ fourth quadrants, thus

x = $sin^{-1}$ ($\frac{1}{3}$ ) ≈ 19.5° ← related acute angle, then

x = 180° + 19.5° = 199.5°

x = 360° - 19.5° = 340.5°

Solution is x = 199.5°, x = 340.5° for 0° < x < 360°