How to Solve Second-Degree Trig Equations Using Pythagorean Identities

In the first three parts of this series, we built up the main tools needed for solving trigonometric equations.

In Part 1, we looked at solving first-degree trigonometric equations on the standard interval. In Part 2, we extended that idea to non-standard intervals and general solutions. In Part 3, we introduced second-degree trigonometric equations that can be solved by treating the trig function like a quadratic.

This fourth part adds one extra step: sometimes the equation does not immediately look like a quadratic in one trig function. Before solving, you first need to use a Pythagorean identity to rewrite the equation in standard quadratic form.

Once that is done, the process is the same as before:

1. Use a Pythagorean identity to rewrite the equation using one trig function

2. Rearrange into standard quadratic form

3. Solve the quadratic by factoring or using the quadratic formula

4. Solve the resulting first-degree trig equations

In this post, we will focus on the standard interval:

0° ≤ θ < 360°

or

0 ≤ θ < 2π

The Main Pythagorean Identities

The three main Pythagorean identities are:

sin²θ + cos²θ = 1

1 + tan²θ = sec²θ

1 + cot²θ = csc²θ

The first identity is the one students usually use most often:

sin²θ + cos²θ = 1

From this, we can rearrange:

sin²θ = 1 − cos²θ

or

cos²θ = 1 − sin²θ

This allows us to change an equation involving both sine and cosine into an equation involving only one trig function.

Example 1: Using sin²θ = 1 − cos²θ

Solve:

2sin²θ + cosθ − 1 = 0, 0° ≤ θ < 360°

At first, this equation has both sine and cosine. That is the problem. We want the equation to contain only one trig function.

Since the equation contains sin²θ and cosθ, we can replace sin²θ with:

1 − cos²θ

So:

2(1 − cos²θ) + cosθ − 1 = 0

Expand:

2 − 2cos²θ + cosθ − 1 = 0

Simplify:

1 − 2cos²θ + cosθ = 0

Rewrite in standard quadratic form:

−2cos²θ + cosθ + 1 = 0

Factor out −1 and then divide both sides by −1 to cancel it out.

−(2cos²θ − cosθ − 1) = 0

2cos²θ − cosθ − 1 = 0

Now factor:

(2cosθ + 1)(cosθ − 1) = 0

So:

2cosθ + 1 = 0

or

cosθ − 1 = 0

Therefore:

cosθ = −1/2

or

cosθ = 1

Now solve each first-degree trig equation.

For:

cosθ = −1/2

cosine is negative in Quadrants II and III, with reference angle 60°. So:

θ = 120°, 240°

For:

cosθ = 1

the solution is:

θ = 0°

Final answer:

θ = 0°, 120°, 240°

Example 2: Using 1 + tan²θ = sec²θ

Solve:

tan²θ − 3secθ + 3 = 0, 0 ≤ θ < 2π

This equation contains tan²θ and secθ. We want to rewrite it using only one trig function.

From the identity:

1 + tan²θ = sec²θ

we get:

tan²θ = sec²θ − 1

Substitute this into the equation:

(sec²θ − 1) − 3secθ + 3 = 0

Simplify:

(sec²θ − 3secθ + 2 = 0

Now factor:

(secθ − 1)(secθ − 2) = 0

So:

secθ = 1

or

secθ = 2

Since secant is the reciprocal of cosine:

secθ = 1 / cosθ

So these become:

cosθ = 1

or

cosθ = 1/2

Now solve each one.

For:

cosθ = 1

we get:

θ = 0

For:

cosθ = 1/2

cosine is positive in Quadrants I and IV, with reference angle 60°. In radians, that gives:

θ = π/3, 5π/3

Final answer:

θ = 0, π/3, 5π/3

Why the Identity Step Matters

The identity step is what changes the problem from a mixed trig equation into a quadratic trig equation.

For example, this equation:

2sin²θ + cosθ − 1 = 0

does not initially look like a normal quadratic because it contains both sine and cosine.

But after replacing:

sin²θ

with:

1 − cos²θ

the equation becomes a quadratic in cosine:

2cos²θ − cosθ − 1 = 0

At that point, the equation can be solved using the same second-degree method from the previous post.

Common Mistakes to Avoid

1. Trying to solve before using the identity

If the equation contains two different trig functions, you often need to rewrite it first. Do not start factoring until the equation is written in terms of one trig function.

2. Forgetting to rearrange into standard form

A quadratic should usually be written as:

ax² + bx + c = 0

For example:

2cos²θ − cosθ − 1 = 0

is ready to factor.

3. Forgetting to solve the trig equations at the end

Factoring the quadratic is not the final answer. If you get:

cosθ = −1/2

you still need to find the actual angle values.

4. Not checking whether the values are possible

For sine and cosine, any value outside the interval:

−1 ≤ y ≤ 1

is impossible.

For example:

cosθ = 3

has no solution.

Summary

Some second-degree trig equations require an extra identity step before the quadratic can be solved.

The full process is:

1. Use a Pythagorean identity to rewrite the equation in one trig function

2. Rearrange into standard quadratic form

3. Solve the quadratic by factoring or the quadratic formula

4. Solve the resulting first-degree trig equations

5. Check that the trig values are possible

The main idea is that the identity does not finish the problem. It simply transforms the equation into a form we already know how to solve.

Once the equation becomes quadratic in one trig function, the rest of the process is the same as solving any second-degree trigonometric equation.

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