Trigonometry Trigonometric Equations 2

Question 1

Solve sin²(x) = 1 for x in [0, 2π).

Accepted Answer

x = π/2, 3π/2

Answer

sin²(x) = 1 means sin(x) = ±1, which occurs at x = π/2 and x = 3π/2 on [0, 2π).

Question 2

Solve 2cos²(x) − cos(x) − 1 = 0 for x in [0, 2π).

Accepted Answer

x = 0, 2π/3, 4π/3

Answer

Factoring gives (2cos(x)+1)(cos(x)−1)=0, so cos(x)=−1/2 (x=2π/3,4π/3) or cos(x)=1 (x=0).

Question 3

Solve sin(x)cos(x) = 1/2 for x in [0, 2π).

Accepted Answer

x = π/4, 5π/4

Answer

Using the identity sin(x)cos(x) = (1/2)sin(2x), the equation becomes sin(2x)=1, giving 2x=π/2+2nπ, so x=π/4, 5π/4.

Question 4

What substitution simplifies solving 2sin²(x) + 3sin(x) − 2 = 0?

Accepted Answer

Let u = sin(x)

Answer

Substituting u = sin(x) converts the equation to the quadratic 2u² + 3u − 2 = 0, which can be factored easily.

Question 5

Solve cos(2x) = cos(x) for x in [0, 2π).

Accepted Answer

x = 0, 2π/3, 4π/3

Answer

Replacing cos(2x) with 2cos²(x)−1 and factoring gives (2cos(x)+1)(cos(x)−1)=0, yielding x=0, 2π/3, 4π/3.

Trigonometry Practice Test