Question

Solve the following equations. (a) 4 cos^2 θ - 4 cos θ +1 = 0 (b) 4 sin^2θ + 2 cos^2θ = 3

Answer 1

The solutions to the given equations are:

For equation (a): θ = arcsin(√(1/2)) or θ = π - arcsin(√(1/2))

For equation (b): θ = arccos(√(1/2)) or θ = -arccos(√(1/2))

Explanation:

To solve equation (a), we can rewrite it using the Pythagorean identity:

4(1 - sin^2θ) - 4cosθ + 1 = 0

Expanding and rearranging, we get:

4 - 4sin^2θ - 4cosθ + 1 = 0

-4sin^2θ - 4cosθ + 5 = 0

Now, let's solve equation (b):

4sin^2θ + 2cos^2θ = 3

Using the Pythagorean identity, we can rewrite it as:

4(1 - cos^2θ) + 2cos^2θ = 3

Expanding and rearranging, we get:

4 - 4cos^2θ + 2cos^2θ = 3

-2cos^2θ + 4 = 3

-2cos^2θ = -1

Dividing by -2, we have:

cos^2θ = 1/2

Taking the square root of both sides, we get:

cosθ = ±√(1/2)

Therefore, the solutions to the equations are:

For equation (a): θ = arcsin(√(1/2)) or θ = π - arcsin(√(1/2))

For equation (b): θ = arccos(√(1/2)) or θ = -arccos(√(1/2))