cos²a + cos²(a+120°) + cos(a-120°) = 3/2
after some of those others you posted, this one is pretty mechanical
cos 120° = -1/2
sin 120° = √3/2
(-1/2)^2 + (cosa * (-1/2) - sina*(√3/2))^2 + (cosa * (-1/2) + sina*(√3/2))^2
1/4 + 1/4 (cos^2a + 2√3 sina*cosa + 3sin^2a) + 1/4 (cos^2a - 2√3 sina*cosa + 3sin^2a)
1/4 (1 + 2cos^2a + 6sin^2a)
1/4 (1 + 2 + 4sin^2a)
3/4 + sin^2a = 3/2
sin^2a = 3/4
sina = ±√3/2
a = pi/3, 2pi/3, 4pi/3, 5pi/3
To solve the equation cos²a + cos²(a+120°) + cos(a-120°) = 3/2, we can use the trigonometric identity:
cos²θ + cos²(θ+120°) + cos(θ-120°) = 1/2 + cosθ
Using this identity, we can rewrite the equation as:
1/2 + cos(a) = 3/2
Subtracting 1/2 from both sides, we get:
cos(a) = 2/2
Simplifying further, we have:
cos(a) = 1
Now, to find the value of angle "a" that satisfies this equality, we can use the inverse cosine function (also known as arccos or cos^(-1)). Taking the inverse cosine of both sides, we get:
a = cos^(-1)(1)
The inverse cosine of 1 is 0 degrees. Therefore, angle "a" is equal to 0 degrees.