# Trigonometry - Solve Equations

## Trigonometry - Solve Equations - How it Works - Video

### Unit Circle

Unit Circle:

The unit circle is vital to answer any exact answer. Do you have to memorize every bit? That is up to. It would be faster if you had it encoded in your brain, but you can memorized just the first quadrant and use symmetry to fill out the unit circle.

### Example 1

Example 1:

Since we have tangent by itself on the left side, we inverse both sides to get phi by itself. We know the angle is sqrt(3). Since we have tangent, we want sqrt(3) to be in the y coordinate in our ordered pair. Remember tangent = y/x. We want this so we don't have to radicalized the denominator.

Here we have two options one in the first quadrant and one in the third quadrant. So phi equals, π /3 or 4π /3. Since the period of tangent is π , every solution is going to be π /3 + π k and other 4π /3 + π k. If k = 1 for π /3 + π k then we get 4π /3. So we can combine the two answers into one, π /3 + π k.

Here we have graphed our equation. You can either graph it as two equations and see where they intersect or set the equation equal to zero, and graph one equation. We set the equation equal to zero, f(φ) = tan(φ) - sqrt(3). So we are looking for the equation crosses the x-axis. The first positive one is π /3. The next one is 4π /3. And the next one is 7π /3. Each time we are adding π .

### Example 2 Part 1

Example 2 Part 1:

Since we have zero on the right side, we can solve for theta. Since both terms involve square numbers, we have a difference of square numbers. Let's factor.

csc4(2θ) - 4 = 0

[csc2(2θ)]2 - 22 = 0

(csc2(2θ) - 2) * (csc2(2θ) + 2)

Rewrote the terms to the be powers.

Factored.

Now we have two factors, so let's set each to zero and solve.

(csc2(2θ) - 2) = 0

csc2(2θ) = 2

csc(2θ) = ± sqrt(2)

Rewrote the terms to the be powers.

Factored.

Let's flip that to sin() = ± sqrt(2)/2.

### Example 2 Part 2

Example 2 Part 2:

Once we have two numbers we substitute the numbers into the formula, cos(u - v) = cos u * cos v + sin u * sin v.

So we get cos(5π /6 - π /4) = cos 5π /6 * cos π /4 - sin 5π /6 * sin π /4, where 5π /6 = u and π /4 = v.

=> cos(5π /6 - π /4) = cos 5π /6 * cos π /4 - sin 5π /6 * sin π /4

=> -sqrt(3) / 2 * sqrt(2) / 2- 1 / 2 * sqrt(2) / 2

=> -sqrt(6) / 4 + sqrt(2) / 4

=> [-sqrt(6) + sqrt(2)] / 4

Substituted u and v.

Use the unit chart to find the corresponding x and y values.

Multiplied numerators and multiplied denominators.

Combined fractions.

So our exact is [-sqrt(6) + sqrt(2)] / 4.

Sine and tangent work the same way.

### Example 2 Part 3

Example 2 Part 3:

We have four options where sin() = ± sqrt(2)/2, one in Q I, one in Q II, one in Q III, and one in Q IV.

Solving for θ for each one:

2θ = π/4

2θ = π/4 + 2πk

θ = π/8 + πk

Q I

Divided by 2.

2θ = 3π/4

2θ = π/4 + 2πk

θ = 3π/8 + πk

Q II

Divided by 2.

2θ = 5π/4

2θ = 5π/4 + 2πk

θ = c + πk

Q III

Divided by 2.

2θ = 7π/4

2θ = 7π/4 + 2πk

θ = 7π/8 + πk

Q IV