Solving Trigonometric Equations Using Pythagorean Identities

Many trig equations don't come pre-packaged with a single function. You'll often see a mix, like sine and cosine, or tangent and secant, sitting in the same equation. That mix is the problem: you can't factor or isolate a variable when two different functions of the same angle are tangled together. The fix is to use a Pythagorean identity to rewrite one of the functions in terms of the other, so the whole equation collapses down to a single trig function.

The three Pythagorean identities you'll lean on are:

\( \sin^2\theta + \cos^2\theta = 1 \)

\( 1 + \tan^2\theta = \sec^2\theta \)

\( 1 + \cot^2\theta = \csc^2\theta \)

Each one can be rearranged to isolate the squared term you need, for example \( \sin^2\theta = 1 - \cos^2\theta \) or \( \tan^2\theta = \sec^2\theta - 1 \).

Once you spot a squared trig term alongside a different trig function of the same angle, follow these steps:

1. Identify which Pythagorean identity connects the two functions in the equation.

2. Substitute to rewrite the squared term so only one trig function remains.

3. Simplify. You'll usually land on a quadratic equation in that single function.

4. Factor (or use the quadratic formula) to solve for the trig function's value.

5. Use a reference angle and the quadrant sign pattern to find every angle solution on the given interval.

6. Check each solution back in the original equation, especially if secant, cosecant, or cotangent were involved, since those have restricted domains.

Solve \( 2\cos^2 x - \sin x - 1 = 0 \) for \( 0 \le x < 2\pi \).

The equation mixes \( \cos^2 x \) and \( \sin x \), so use \( \cos^2 x = 1 - \sin^2 x \):

\( 2(1-\sin^2 x) - \sin x - 1 = 0 \)

\( 2 - 2\sin^2 x - \sin x - 1 = 0 \)

\( -2\sin^2 x - \sin x + 1 = 0 \)

Multiply through by \(-1\) so the leading coefficient is positive:

\( 2\sin^2 x + \sin x - 1 = 0 \)

This factors nicely:

\( (2\sin x - 1)(\sin x + 1) = 0 \)

So \( \sin x = \frac{1}{2} \) or \( \sin x = -1 \).

For \( \sin x = \frac{1}{2} \), the reference angle is \( \frac{\pi}{6} \), and sine is positive in quadrants I and II, giving \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \). For \( \sin x = -1 \), the only solution on this interval is \( x = \frac{3\pi}{2} \).

The graph below shows \( y = 2\cos^2 x - \sin x - 1 \) on this interval. Every point where the curve crosses the x-axis is one of the solutions found above.

So the full solution set is \( x = \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{3\pi}{2} \).

Solve \( \tan^2 x - \sec x - 1 = 0 \) for \( 0 \le x < 2\pi \).

Here the mix is tangent and secant, so use \( \tan^2 x = \sec^2 x - 1 \):

\( (\sec^2 x - 1) - \sec x - 1 = 0 \)

\( \sec^2 x - \sec x - 2 = 0 \)

Factoring gives:

\( (\sec x - 2)(\sec x + 1) = 0 \)

So \( \sec x = 2 \) or \( \sec x = -1 \), which means \( \cos x = \frac{1}{2} \) or \( \cos x = -1 \).

For \( \cos x = \frac{1}{2} \), cosine is positive in quadrants I and IV, so \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \). For \( \cos x = -1 \), the solution is \( x = \pi \).

Solution set: \( x = \frac{\pi}{3}, \ \pi, \ \frac{5\pi}{3} \).

Watch for these traps when solving this type of equation:

• Forgetting that squaring or substituting can introduce values outside the function's actual range, like solving for \( \cos x = 2 \), which has no real solution and should be discarded.

• Losing track of the sign when substituting the identity, especially when rearranging \( \sin^2\theta = 1 - \cos^2\theta \).

• Mixing up which quadrants give a positive or negative value for the function you've solved for. Reviewing the ASTC rule keeps the sign pattern straight.

• Stopping after finding the value of the trig function instead of converting it back into every valid angle on the interval.

Once substituting a Pythagorean identity feels comfortable, the next step is equations built from a double angle, where you'll combine this same factoring approach with double angle identities to reduce the equation to a single angle before solving.

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