I guess you mean cos (3θ -120°) = √ (3 /2 ) = sqrt 1.5
nope no way you must mean (1/2) sqrt 3 = no, that is also >1
The leg is never longer than the hypotenuse.
Solve the equation cos (3θ -120°) = √(3 /2
3 answers
whoops (1/2) sqrt 3 = 0.866 which is fine
so the angle in the brackets might be 30 degrees or -30 degrees
cos (3(30)-120) = cos(-30) = .866 so 30 degrees works
so the angle in the brackets might be 30 degrees or -30 degrees
cos (3(30)-120) = cos(-30) = .866 so 30 degrees works
Since you are showing degrees, I will continue in degrees.
cos (3θ -120°) = √(3 /2)
I know that cos 30° = √3/2 , from my special angles list
and since the cosine is positive in I and IV
cos 330° = √3/2
so 3θ -120° = 30° or 3θ -120° = 330°
3θ = 120° + 30° or 3θ = 120° + 330°
3θ = 150° or 3θ = 450°
θ = 50° or θ = 150°
since the cos 3θ is 360°/3 = 120°
more solutions can be obtained by adding and/or subtracting multiples
of 120°
You didn't state a domain, so assuming 0 ≤ θ ≤ 360°
θ = 50°, 150°, 170°, 270°, 290°
cos (3θ -120°) = √(3 /2)
I know that cos 30° = √3/2 , from my special angles list
and since the cosine is positive in I and IV
cos 330° = √3/2
so 3θ -120° = 30° or 3θ -120° = 330°
3θ = 120° + 30° or 3θ = 120° + 330°
3θ = 150° or 3θ = 450°
θ = 50° or θ = 150°
since the cos 3θ is 360°/3 = 120°
more solutions can be obtained by adding and/or subtracting multiples
of 120°
You didn't state a domain, so assuming 0 ≤ θ ≤ 360°
θ = 50°, 150°, 170°, 270°, 290°