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7.4PureStretch

Solving Trigonometric Equations

Harder trig equations require an identity — an addition or double-angle formula — to reduce them to a single trig function before solving. Always check the interval.

28 min Trigonometry and Modelling
Edexcel A level Maths: 7.4 Solving Trigonometric EquationsCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Use double-angle formulae to simplify equations
  • Reduce to a single trig function
  • Solve the resulting quadratic or linear equation
  • List all solutions in the interval
1

Strategy

Replace double or compound angles using an identity so the whole equation is in one function and one angle. Then solve as a standard equation and apply the interval.

Example 1Solve cos⁡2θ+sin⁡θ=0\cos 2\theta + \sin\theta = 0 for 0≤θ<2π0 \le \theta < 2\pi.
1Use cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta: 1−2sin⁡2θ+sin⁡θ=01 - 2\sin^2\theta + \sin\theta = 0.
22sin⁡2θ−sin⁡θ−1=0⇒(2sin⁡θ+1)(sin⁡θ−1)=02\sin^2\theta - \sin\theta - 1 = 0 \Rightarrow (2\sin\theta + 1)(\sin\theta - 1) = 0.
3sin⁡θ=−12\sin\theta = -\tfrac12 or sin⁡θ=1\sin\theta = 1.
Answer
θ=π2,7π6,11π6\theta = \tfrac{\pi}{2}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}

Tip — Choose the cos 2θ form that leaves only the function already in the equation.

Formula recap

cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta
Reduce to sine.
cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1
Reduce to cosine.
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta
Factor a common term.

Common mistakes to avoid

Dividing through by sin θ and losing solutions.
Factor instead — dividing can drop valid roots.
Forgetting solutions in the interval.
Use symmetry and period to list every solution.

Key takeaways

  • Use an identity to reduce to one function and one angle.
  • Form a quadratic and factor (don’t divide out).
  • List all solutions in the given interval.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. To solve cos⁡2θ+sin⁡θ=0\cos 2\theta + \sin\theta = 0, replace cos 2θ with:

    • 1−2sin⁡2θ1 - 2\sin^2\theta
    • 2cos⁡2θ−12\cos^2\theta - 1
    • 2sin⁡θcos⁡θ2\sin\theta\cos\theta
    • cos⁡2θ\cos^2\theta
    Show answer and explanation

    Answer: 1−2sin⁡2θ1 - 2\sin^2\theta

    Leaves only sine.

  2. When an equation has sin θ as a common factor, you should:

    • factor
    • divide by sin θ
    • square it
    • ignore it
    Show answer and explanation

    Answer: factor

    Dividing loses roots.

  3. 2sin⁡2θ−sin⁡θ−1=02\sin^2\theta - \sin\theta - 1 = 0 factors as:

    • (2sin⁡θ+1)(sin⁡θ−1)(2\sin\theta + 1)(\sin\theta - 1)
    • (2sin⁡θ−1)(sin⁡θ+1)(2\sin\theta - 1)(\sin\theta + 1)
    • (sin⁡θ−1)2(\sin\theta - 1)^2
    • (2sin⁡θ+1)2(2\sin\theta + 1)^2
    Show answer and explanation

    Answer: (2sin⁡θ+1)(sin⁡θ−1)(2\sin\theta + 1)(\sin\theta - 1)

    Check by expanding.

  4. sin⁡θ=1\sin\theta = 1 on [0,2π)[0, 2\pi) gives:

    • π2\tfrac{\pi}{2}
    • 00
    • π\pi
    • 3π2\tfrac{3\pi}{2}
    Show answer and explanation

    Answer: π2\tfrac{\pi}{2}

    Maximum of sine.

  5. The goal when simplifying a trig equation is to get:

    • one function, one angle
    • two functions
    • a cubic
    • degrees only
    Show answer and explanation

    Answer: one function, one angle

    Then it solves directly.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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