# Trigonometric Equation

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

## Example 1

### Example

### Solution

cos^{2} x = 1 - sin^{2} x

Pythagorean Identity

1 + 1 = 2

Multiply -1 on both sides.

Then sin^{2} x + sin x - 2 = 0.

Think sin x as a variable

and factor sin^{2} 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 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:

1^{2} + [height]^{2} = 2^{2}.

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.