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Solving trigonometric equations using double-angle identities

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Solving Trigonometric Equations Using Double-Angle Identities

This lesson shows how to rewrite trigonometric equations containing a double angle, like sin 2x or cos 2x, into single-angle equations using the double-angle identities, then solve them with factoring and the unit circle.

Why double-angle identities show up in trig equations

Many trigonometric equations don't come in a nice, single-angle form. Instead you'll see something like \(\sin(2x) = \sin(x)\) or \(\cos(2x) + \cos(x) = 0\), where the angle \(2x\) is mixed in with \(x\). Before you can solve for \(x\), you need to rewrite every \(2x\) term using the double angle identities so that the whole equation is written in terms of a single angle. Once that's done, solving looks a lot like the methods used for solving second degree trigonometric equations: factor, set each factor to zero, and solve.

The identities you'll substitute

These are the three double-angle identities that come up again and again:

\(\sin(2\theta) = 2\sin\theta\cos\theta\)

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

\(\tan(2\theta) = \dfrac{2\tan\theta}{1-\tan^2\theta}\)

Notice that cosine has three equivalent versions. Picking the right one is often the key move: if the equation already has a lone \(\cos x\) or \(\sin x\) term sitting next to \(\cos 2x\), choose the version of the identity that leaves only that same function behind, so the whole equation ends up in one trig function.

A step-by-step strategy

  1. Look for any \(2x\) (or \(2\theta\)) term and replace it using the matching double-angle identity.
  2. Simplify so the equation contains only one trig function, or factor it into a product of simpler expressions.
  3. Set each factor equal to zero and solve each resulting basic equation.
  4. Use the ASTC rule and a reference angle to find every solution in the required interval, not just one.
  5. Check each solution against the original equation and the given domain.

Worked example 1: \(\sin(2x) = \sin(x)\)

Solve for \(0 \le x < 2\pi\).

Step 1: Substitute the identity for \(\sin(2x)\):

\(2\sin x\cos x = \sin x\)

Step 2: Move everything to one side and factor. Do not divide both sides by \(\sin x\), since that can throw away solutions.

\(2\sin x\cos x - \sin x = 0\)

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

Step 3: Set each factor to zero.

\(\sin x = 0\) or \(\cos x = \dfrac{1}{2}\)

Step 4: Solve each piece on \([0, 2\pi)\). For \(\sin x = 0\), \(x = 0\) or \(x = \pi\). For \(\cos x = \frac{1}{2}\), the reference angle is \(\frac{\pi}{3}\), and cosine is positive in quadrants I and IV, giving \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\).

The full solution set is \(x = 0, \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3}\).

Graph of y equals sin of 2x minus sin of x from 0 to 2 pi, showing four x-intercepts Plot of y = sin(2*x) - sin(x) for x in [0, 6.2832] 0 1 2 3 4 5 6 -2 -1 0 1 2 x y x = 0 x = pi/3 x = pi x = 5pi/3
Graph of \(y = \sin(2x) - \sin(x)\), with the x-intercepts marking the solutions found above.

Worked example 2: \(\cos(2x) + \cos(x) = 0\)

Solve for \(0 \le x < 2\pi\).

Here the equation has a lone \(\cos x\) term, so choose the version of the cosine double-angle identity written entirely in cosine: \(\cos(2x) = 2\cos^2 x - 1\).

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

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

This is a quadratic in \(\cos x\). Factoring gives:

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

So \(\cos x = \dfrac{1}{2}\) or \(\cos x = -1\). Using the reference angle and the ASTC rule for the first equation, and recalling the exact value for the second, the solutions on \([0, 2\pi)\) are \(x = \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3}\).

Graph of y equals cos of 2x plus cos of x from 0 to 2 pi, showing three x-intercepts Plot of y = cos(2*x) + cos(x) for x in [0, 6.2832] 0 1 2 3 4 5 6 -1 0 1 2 x y x = pi/3 x = pi x = 5pi/3
Graph of \(y = \cos(2x) + \cos(x)\), showing the same three solutions as x-intercepts.

Common pitfalls to avoid

  • Dividing out a common factor. In example 1, dividing both sides by \(\sin x\) instead of factoring would silently erase the solutions \(x = 0\) and \(x = \pi\).
  • Picking the wrong cosine identity. Using \(\cos(2x) = \cos^2 x - \sin^2 x\) in example 2 would leave a mix of sine and cosine that's harder to factor. Choosing \(2\cos^2 x - 1\) matched the lone \(\cos x\) already in the equation.
  • Forgetting the full domain. Because \(2x\) moves through the interval twice as fast as \(x\), double-angle equations often have more solutions than a same-looking single-angle equation. Always check every quadrant using the reference angle.
  • Losing track of exact values. Many final answers rely on knowing exact trig ratios; reviewing how to find the exact value of trigonometric ratios makes these steps much faster.

Practice tip

Before tackling a double-angle equation, make sure you're comfortable with the identities themselves and with solving first degree trigonometric equations, since every double-angle problem eventually reduces to that same kind of basic equation once the substitution and factoring are done.

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