What are the solutions for 3cos2x=-3 if 0≤ x ≤ 2pi

Answer 1

That is Trigonometry problem

Trigonometry isn't Calculus

3 cos 2x = - 3

Divide both sides by 3

cos 2x = - 1

2 x = cos⁻¹ ( - 1 )

2 x = arccos ( - 1 )

arccos ( - 1 ) = π

The period of cosine function is 2 π so:

2 x = π + 2 π n

Divide both sides by 2

x = π / 2 + π n

n = 0 , ±1 , ±2 , ±3...

For 0 ≤ x ≤ 2 π

x = π / 2 + π ∙ 0 = π / 2 + 0 = π / 2

and

x = π / 2 + π ∙ 1 = π / 2 + π = π / 2 + 2 π / 2 = 3 π / 2