Trigonometric Equation

How to solve the given trigonometric equation by using the trigonometric identities: 2 examples and their solutions.

Example 1

Example

Solution

cos2 x = 1 - sin2 x

Pythagorean Identity

1 + 1 = 2

Multiply -1 on both sides.
Then sin2 x + sin x - 2 = 0.

Think sin x as a variable
and factor sin2 x + sin x - 2 = 0.

Find a pair of numbers
whose product is the constant term -2
and whose sum is the coefficient of the middle term +1.

-1⋅2 = -2
-1 + 2 = +1

So (sin x - 1)(sin x + 2) = 0.

Factor a Quadratic Trinomial

Solve the quadratic equation for each case.

Case 1: sin x - 1 = 0

sin x = 1

Solve sin x = 1.

Sine: Equation

First find one of the solutions.
θ = π/2

Sine Values of Commonly Used Angles

Then the general solution of sin x = 1 is
x = nπ + (-1)n⋅[π/2].

Find the x values
that are in (0 ≤ x ≤ 2π).

n = 0

x = 0⋅π + (-1)0⋅[π/2]
= π/2

This is in (0 ≤ x ≤ 2π).

n = 1

x = 1⋅π + (-1)1⋅[π/2]
= π/2

This is in (0 ≤ x ≤ 2π).

n = 2

x = 2⋅π + (-1)2⋅[π/2]
= 2π + π/2

This is greater than 2π.
So this is not in (0 ≤ x ≤ 2π).

So, for case 1,
x = π/2.

Case 2: sin x + 2 = 0

sin x = -2

sin x = -2

But -1 ≤ sin x ≤ 1.
(y = sin x moves between -1 and 1.)

Sine: Graph

So sin x = -2 has no solution.

So, for case 2,
there's no solution.

For case 1,
x = π/2.

For case 2,
there's no solution.

So the solution of (sin x - 1)(sin x + 2) = 0 is
x = π/2.

So x = π/2 is the answer.

Example 2

Example

Solution

Move sin x to the left side.

sin 2x = 2 sin x cos x

sin 2A

2 sin x cos x - sin x = 0

The common factor is sin x.

So sin x (2 cos x - 1) = 0.

Solve the equation for each case.

Case 1: sin x = 0

Solve sin x = 0.

First find one of the solutions.
θ = 0

Sine Values of Commonly Used Angles

Then the general solution of sin x = 0 is
x = nπ + (-1)n⋅0.

Find the x values
that are in (0 ≤ x ≤ 2π).

n = 0

x = 0⋅π + (-1)0⋅0
= 0

This is in (0 ≤ x ≤ 2π).

n = 1

x = 1⋅π + (-1)1⋅0
= π

This is in (0 ≤ x ≤ 2π).

n = 2

x = 2⋅π + (-1)2⋅0
= 2π

This is in (0 ≤ x ≤ 2π).

If n gets bigger,
then x will not be in (0 ≤ x ≤ 2π).

So stop finding more solutions.

So, for case 1,
x = 0, π, 2π.

Case 2: 2 cos x - 1 = 0

cos x = 1/2

Solve cos x = 1/2.

Cosine: Equation

First find one of the solution.

Cosine is CAH:
Cosine,
Adjacent side (1),
Hypotenuse (2).

So draw a right triangle
whose adjacent side is 1
and whose hypotenuse is 2.

Find the height
by using the Pythagorean theorem:
12 + [height]2 = 22.

Then the height is √3.

This is a right triangle
whose sides are 1, √3, and 2.

So this is a 30-60-90 triangle.

So the central angle is, 60º, π/3.

Radian Measure

θ = π/3

Then the general solution of cos x = 1/2 is
x = 2nπ ± π/3.

Find the x values
that are in (0 ≤ x ≤ 2π).

n = 0

x = 2⋅0⋅π ± π/3
= ± π/3

π/3 is in (0 ≤ x ≤ 2π).
So x = π/3.

n = 1

x = 2⋅1⋅π ± π/3
= 7π/3, 5π/3

5π/3 is in (0 ≤ x ≤ 2π).
So x = 5π/3.

So, for case 2,
x = π/3, 5π/3.

For case 1,
x = 0, π, 2π.

For case 2,
x = π/3, 5π/3.

So the solution of sin x (2 cos x - 1) = 0 is
x = 0, π/3, π, 5π/3, 2π.

So x = 0, π/3, π, 5π/3, 2π.