To solve the equation, we can use the double-angle formula for cosine:
cos(2θ) = 2cos^2(θ) - 1
Let's denote x as θ to make it easier. So, the equation becomes:
cos(0.5x) + cos(0.25x) = 0
Using the double-angle formula for cosine, we get:
2cos^2(0.25x) - 1 + 2cos^2(0.125x) - 1 = 0
2cos^2(0.25x) + 2cos^2(0.125x) = 2
cos^2(0.25x) + cos^2(0.125x) = 1
But since cos^2(θ) + sin^2(θ) = 1, we can rewrite the equation as:
sin^2(0.25x) + cos^2(0.125x) = 1
Which means:
sin(0.25x) = 1 and cos(0.125x) = 0
Solving for x:
0.25x = π/2
x = 2π
To solve the equation, we can use the double-angle formula for cosine:
cos(2θ) = 2cos^2(θ) - 1
Let's denote x as θ to make it easier. So, the equation becomes:
cos(0.5x) + cos(0.25x) = 0
Using the double-angle formula for cosine, we get:
2cos^2(0.25x) - 1 + 2cos^2(0.125x) - 1 = 0
2cos^2(0.25x) + 2cos^2(0.125x) = 2
cos^2(0.25x) + cos^2(0.125x) = 1
But since cos^2(θ) + sin^2(θ) = 1, we can rewrite the equation as:
sin^2(0.25x) + cos^2(0.125x) = 1
Which means:
sin(0.25x) = 1 and cos(0.125x) = 0
Solving for x:
0.25x = π/2
x = 2π