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Concept 08 / 11Trigonometry
Trigonometric Equations: General Solutions and Methods
Principal and general solutions of sin x = sin a, cos x = cos a and tan x = tan a, equations reducing to quadratics, a sin x + b cos x = c, factorisation and sum-to-product, squaring and extraneous roots, and solutions in a given interval.
Figure 1
A trigonometric equation usually has infinitely many solutions because the functions are periodic. The solutions in [0,2π) (or the one in the principal range) are principal solutions; a formula describing all of them is the general solution.
The three basic general solutions (n∈Z):
sinx=sinα⟺x=nπ+(−1)nα.
cosx=cosα⟺x=2nπ±α.
tanx=tanα⟺x=nπ+α.
They come from symmetry and periodicity: sin takes each value twice per period (at α and π−α), cos at ±α, and tan once per period π. The figure shows the solutions of sinx=21 generated this way. Squared versions combine both signs: sin2x=sin2α, cos2x=cos2α and tan2x=tan2α all have the solution x=nπ±α.
Special cases: sinx=0⟺x=nπ; cosx=0⟺x=(2n+1)2π; sinx=1⟺x=2nπ+2π; cosx=1⟺x=2nπ; cosx=−1⟺x=(2n+1)π.
Methods.
Reduce to one function: use identities to express the equation in a single ratio, often giving a quadratic, e.g. 2cos2x+3sinx=0 becomes 2−2sin2x+3sinx=0. Reject roots outside [−1,1].
Factorise: move everything to one side and factor, often after sum-to-product. Never divide both sides by an expression that could be zero (such as cosx) without separately checking when it is zero; dividing loses those solutions.
asinx+bcosx=c: divide by a2+b2 to get sin(x+α)=a2+b2c; solvable iff ∣c∣≤a2+b2.
Homogeneous equations in sinx and cosx (every term of the same degree): divide by the appropriate power of cosx to get an equation in tanx, after checking whether cosx=0 is a solution.
Substitutions: t=sinx+cosx turns sinxcosx into 2t2−1, handling equations symmetric in sin and cos.
Boundedness: equations like sinx+cosy=2 force each term to its maximum, sinx=1 and cosy=1. Similarly sin2x+cos4x=1 can be analysed by comparing with sin2x+cos2x=1.
Extraneous roots. Squaring both sides introduces solutions of the equation with the opposite sign. After squaring, substitute each candidate back into the original equation. Similarly, check that no solution makes a term like tanx or secx undefined.
Solutions in an interval. Write the general solution, then list the values of n for which it lies in the interval. Graphs help count solutions without listing them.
Key Takeaways:
- sinx=sinα: nπ+(−1)nα; cosx=cosα: 2nπ±α; tanx=tanα: nπ+α.
- Squared forms: nπ±α.
- Do not divide by possibly-zero factors; factorise instead.
- asinx+bcosx=c needs ∣c∣≤a2+b2.
- After squaring, verify every root.
- Boundedness can force each term to an extreme value.
- Pitfall: writing sinx=sinα⇒x=2nπ+α and missing the π−α family.
Brain-Teaser 1: Solve 3cosx+sinx=2.
Solution: Divide by 2: 23cosx+21sinx=21, i.e. cos(x−6π)=cos4π. So x−6π=2nπ±4π, giving x=2nπ+125π or x=2nπ−12π. Converting to a single cosine avoids squaring and its extraneous roots.
Brain-Teaser 2: Solve tanx+secx=3 for x∈[0,2π].
Solution: Write cosxsinx+1=3, so 3cosx−sinx=1, i.e. cos(x+6π)=21. Then x+6π=2nπ±3π: x=2nπ+6π or x=2nπ−2π. The second family makes cosx=0, where tan and sec are undefined, so it must be rejected. In [0,2π] the only solution is x=6π. Multiplying through by cosx silently admitted the invalid roots.
Brain-Teaser 3: Find all x with sin4x+cos4x=85.
Solution: sin4x+cos4x=1−2sin2xcos2x=1−21sin22x. So sin22x=43, i.e. sin22x=sin23π, giving 2x=nπ±3π and x=2nπ±6π. Reducing the quartic expression to a single squared sine makes the squared-form general solution applicable directly.
