Answer to Question #126026 in Trigonometry for Habeeb goat

Question

Find all solutions in [0,2π) to the equation

2sin²(A)=cos(A)+1

SHOW ALL WORK

Answer 1

Solution.

2sin²(A) = cos(A) + 1.

We know, that sin²x + cos²x = 1 (1)

By subtracting cos²x from both sides of equation (1) we will receive:

sin²x = 1 - cos²x (2)

Now we use equation (2) for our condition of the task, we will receive:

2(1 - cos²(A)) = cos(A) + 1,

Now we open parentheses and subtract cos(A) + 1 from both sides of the equation:

2 - 2cos²(A) - cos(A) - 1 = 0,

-2cos²(A) - cos(A) + 1 = 0

Multiplying both sides by (-1) we will get:

2cos²(A) + cos(A) - 1 = 0.

Let cos(A) = x, then we will get:

2x² + x - 1 = 0.

Here we have quadratic equation. To find x we will use next formulas:

if ax² + bx + c = 0, then

D = b² - 4ac,

$x_1=\frac{\left(- b - \sqrt{D}\right)}{2a}$

$x_2=\frac{\left(-b+\sqrt{D}\right)}{2a}$

Let's find x from our equation:

2x² + x - 1 = 0

D = 1² - 4 * 2 *(-1) = 1 + 8 = 9 = 3²,

x₁ = (-1 - 3) / 2 * 2 = -4 / 4 = -1,

x₂ = (-1 + 3) / 2 * 2 = 2 / 4 = 1/2.

Now we use our substitution cos(A) = x and we will receive:

cos(A₁) = -1 and cos(A₂) = 1/2.

A₁ = cos^-1(-1) = π,

A₂ = cos^-1(1/2), in [0,2π) we have two values of A: A₂ = π/3, A₃ = 5π/3.

Answer: A₁ = π, A₂ = π/3, A₃ = 5π/3.