How do I solve cos θ + 2 sin θ = 2?

This is an equation in trigonometry that involves the cosine and sine functions of an angle θ. The equation is cos θ + 2 sin θ = 2. Solving for θ requires knowledge of trigonometric identities and algebraic manipulation. If you have any questions about how to solve this equation, feel free to ask!

How do I solve cos θ + 2 sin θ = 2?

To solve the equation cos θ + 2 sin θ = 2, we can use the fact that sin^2 θ + cos^2 θ = 1.

First, we can square both sides of the equation to get rid of the square root:

(cos θ + 2 sin θ)^2 = 2^2

Expanding the left side of the equation gives:

cos^2 θ + 4 sin θ cos θ + 4 sin^2 θ = 4

Next, we can use the identity sin^2 θ + cos^2 θ = 1 to substitute for cos^2 θ:

1 – sin^2 θ + 4 sin θ cos θ + 4 sin^2 θ = 4

Simplifying the equation gives:

5 sin^2 θ + 4 sin θ cos θ – 3 = 0

We can then use the quadratic formula to solve for sin θ:

sin θ = (-b ± sqrt(b^2 – 4ac)) / 2a

where a = 5, b = 4 cos θ, and c = -3.

Substituting these values into the formula and simplifying gives:

sin θ = (-2 ± sqrt(16 cos^2 θ + 60)) / 10

We can then solve for cos θ using the identity cos^2 θ + sin^2 θ = 1:

cos θ = ± sqrt(1 – sin^2 θ)

Substituting the expression for sin θ into this formula gives:

cos θ = ± sqrt(1 – (-2 ± sqrt(16 cos^2 θ + 60)) / 10)^2

This is a bit messy, but we can simplify it by making the substitution u = cos^2 θ:

cos θ = ± sqrt(1 – (-2 ± 2 sqrt(4u + 15)) / 10)^2

cos θ = ± sqrt(1 + (-2 ± 2 sqrt(4u + 15)) / 10)^2

cos θ = ± (1 + (-2 ± 2 sqrt(4u + 15)) / 10)

We can then use the fact that cos^2 θ + sin^2 θ = 1 to solve for sin θ:

sin θ = ± sqrt(1 – cos^2 θ)

Substituting the expression for cos θ into this formula gives:

sin θ = ± sqrt(1 – (1 + (-2 ± 2 sqrt(4u + 15)) / 10)^2)

This gives us the solutions for sin θ and cos θ in terms of u. We can then solve for u using the original equation:

cos θ + 2 sin θ = 2

Substituting the expressions for sin θ and cos θ in terms of u gives:

sqrt(u) ± 2 sqrt(1 – u) = 2

Squaring both sides and simplifying gives:

5u^2 – 8u + 3 = 0

This is a quadratic equation that can be factored as:

(5u – 3)(u – 1) = 0

Therefore, the solutions for u are u = 3/5 and u = 1.

Substituting these values back into the expressions for sin θ and cos θ gives:

sin θ = ± sqrt(1 – cos^2 θ) = ± sqrt(16/25) = ± 4/5

cos θ = ± sqrt(1 – sin^2 θ) = ± sqrt(9/25) = ± 3/5

Therefore, the solutions