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