Trigonometric Equations
When a ratio is fixed but the angle is unknown — and why one equation has infinitely many solutions packed into a single formula
- The difference between a trigonometric identity and a trigonometric equation.
- Principal versus general solutions, and why the answer is a whole family.
- The three master formulae for \(\sin\theta=\sin\alpha\), \(\cos\theta=\cos\alpha\), \(\tan\theta=\tan\alpha\).
- The squared forms \(\sin^2\theta=\sin^2\alpha\) etc., and the special values \(0,\pm1\).
- The four working methods: factorise, reduce to one ratio, the \(a\sin\theta+b\cos\theta=c\) form, and squaring.
- How to discard extraneous roots introduced by squaring or by domain restrictions.
Equation versus Identity
An identity such as \(\sin^2\theta+\cos^2\theta=1\) is true for every value of the angle. A trigonometric equation such as \(2\sin\theta=1\) is true only for particular values. Chapter 11 built the identities; this chapter asks the opposite question — given a relation, which angles satisfy it?
Because the six ratios are periodic, an equation that has one solution automatically has infinitely many. Our job is not to list them but to capture them all in one compact general solution.
Principal & General Solutions
The principal solutions are those lying in \([0,2\pi)\) — the first lap around the circle. The general solution then adds every full turn back in, using an integer parameter \(n\in\mathbb{Z}\).
Throughout this chapter \(n\) is any integer, \(n\in\mathbb{Z}\), and \(\alpha\) denotes the principal angle satisfying the equation. Different forms of the general solution can look different yet describe the same set — always acceptable if equivalent.
The Sine Equation
If \(\sin\theta=\sin\alpha\), the two angles either coincide (up to full turns) or are supplementary. Both cases fold into a single alternating formula.
Even \(n\) gives \(2k\pi+\alpha\); odd \(n\) gives \((2k+1)\pi-\alpha\), the supplementary branch. Here \(\alpha\) is taken in \(\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\).
The Cosine Equation
Cosine is an even function, so \(\cos\theta=\cos\alpha\) is satisfied by \(+\alpha\) and \(-\alpha\) alike — the two angles are reflections across the \(x\)-axis.
The \(\pm\) captures both the angle and its mirror image; \(\alpha\) is taken in \([0,\pi]\).
The Tangent Equation
Tangent has the shorter period \(\pi\), so its general solution is the simplest of the three.
Valid provided \(\theta\ne(2k+1)\tfrac{\pi}{2}\), where tangent is undefined; \(\alpha\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\).
Squared Equations
When an equation reduces to a squared ratio — common after applying a Pythagorean or double-angle identity — all three collapse to one neat form.
Special Values
The values \(0,1,-1\) occur so often that their solutions are worth memorising directly rather than rederiving each time.
| Equation | General solution |
|---|---|
| \(\sin\theta=0\) | \(\theta=n\pi\) |
| \(\sin\theta=1\) | \(\theta=2n\pi+\dfrac{\pi}{2}\) |
| \(\sin\theta=-1\) | \(\theta=2n\pi-\dfrac{\pi}{2}\) |
| \(\cos\theta=0\) | \(\theta=(2n+1)\dfrac{\pi}{2}\) |
| \(\cos\theta=1\) | \(\theta=2n\pi\) |
| \(\cos\theta=-1\) | \(\theta=(2n+1)\pi\) |
| \(\tan\theta=0\) | \(\theta=n\pi\) |
Method 1 — Factorisation
Polynomial equations in a single ratio are solved exactly as ordinary algebra: bring everything to one side, factor, and set each factor to zero.
For \(2\sin^2\theta-3\sin\theta+1=0\), treat \(s=\sin\theta\) as the unknown: \((2s-1)(s-1)=0\) gives \(\sin\theta=\tfrac12\) or \(\sin\theta=1\), then apply the master formulae to each.
Method 2 — Reduce to One Ratio
If two different ratios appear, use a Chapter 11 identity to express everything through a single ratio. A Pythagorean identity converts \(\cos^2\theta\) into \(1-\sin^2\theta\); sum-to-product collapses sums of sines or cosines into a single product.
Replacing \(\cos^2\theta\to1-\sin^2\theta\) turns a mixed equation into a quadratic in \(\sin\theta\). Grouping like-weighted terms and applying \(\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2}\) exposes a common factor that splits the problem cleanly.
Method 3 — \(a\sin\theta+b\cos\theta=c\)
The linear combination from Chapter 11 reappears, now as an equation. Write the left side as a single sine and the rest is the cosine equation.
An equation like \(3\sin\theta+4\cos\theta=8\) has no solution, since the left side never exceeds \(\sqrt{3^2+4^2}=5\). Always compare \(|c|\) with \(\sqrt{a^2+b^2}\) first.
Squaring & Lost or Extra Roots
Squaring both sides — or multiplying out a denominator — can introduce angles that do not satisfy the original equation. Cancelling a common factor can instead lose genuine roots.
Never cancel a trigonometric factor; set it to zero instead — cancelling \(\cos\theta\) in \(\sin\theta\cos\theta=\cos\theta\) would silently discard the family \(\cos\theta=0\). And after any squaring, substitute each candidate back; reject those that fail the original, including any that make a \(\tan\), \(\sec\), \(\csc\) or \(\cot\) undefined.
Putting It to Work
Problem. Solve \(\sin\theta=\dfrac{\sqrt3}{2}\).
Solution. The principal angle is \(\alpha=\tfrac{\pi}{3}\), so by the sine formula:
Problem. Solve \(2\sin^{2}\theta-3\sin\theta+1=0\).
Solution. Factor as \((2\sin\theta-1)(\sin\theta-1)=0\), giving \(\sin\theta=\tfrac12\) or \(\sin\theta=1\):
Problem. Solve \(\sin\theta+\sin3\theta+\sin5\theta=0\).
Solution. Pair the outer terms: \(\sin\theta+\sin5\theta=2\sin3\theta\cos2\theta\), so the equation becomes \(\sin3\theta\,(2\cos2\theta+1)=0\):
Problem. Solve \(\sqrt3\cos\theta+\sin\theta=1\).
Solution. Here \(R=2\); divide by \(2\) to read off \(\cos\!\left(\theta-\tfrac{\pi}{6}\right)=\tfrac12\):
Problem. Solve \(\sin2\theta=\cos\theta\).
Solution. Write \(2\sin\theta\cos\theta-\cos\theta=0\), i.e. \(\cos\theta\,(2\sin\theta-1)=0\). Keep both factors:
Problem. Find the solutions of \(\tan\theta+\sec\theta=2\cos\theta\) in \([0,2\pi)\).
Solution. Multiply by \(\cos\theta\): \(\sin\theta+1=2\cos^2\theta=2(1-\sin^2\theta)\), giving \(2\sin^2\theta+\sin\theta-1=0\), so \((2\sin\theta-1)(\sin\theta+1)=0\).
The root \(\sin\theta=-1\) makes \(\sec\theta\) undefined, so only two solutions survive.
Chapter Summary
\(\sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha\).
\(\cos\theta=\cos\alpha\Rightarrow\theta=2n\pi\pm\alpha\).
\(\tan\theta=\tan\alpha\Rightarrow\theta=n\pi+\alpha\).
All three squared forms \(\Rightarrow\theta=n\pi\pm\alpha\).
Factorise; reduce to one ratio; \(a\sin\theta+b\cos\theta=c\); square.
Check \(|k|\le1\); never cancel a factor; back-check after squaring.
Problems
Give the general solution unless an interval is specified. Difficulty rises down the list.
- Solve \(\sin\theta=-\tfrac12\).
- Solve \(\sqrt3\tan\theta=1\).
- Solve \(2\cos^{2}\theta=1\) (write the answer as \(n\pi\pm\alpha\)).
- Solve \(4\sin^{2}\theta-8\sin\theta+3=0\).
- Solve \(\sin2\theta+\cos\theta=0\).
- Solve \(\tan\theta+\tan2\theta+\sqrt3\,\tan\theta\tan2\theta=\sqrt3\).
- Solve \(\cos\theta+\cos3\theta+\cos5\theta+\cos7\theta=0\).
- Solve \(\sin\theta+\sqrt3\cos\theta=1\).
- Find the general solution of \(\sin\theta+\sin2\theta+\sin3\theta=0\).
- Solve \(3\cos^{2}\theta-2\sqrt3\sin\theta\cos\theta-3\sin^{2}\theta=0\).
- Solve \(\sec\theta-1=(\sqrt2-1)\tan\theta\).
- Find the number of solutions of \(\tan\theta+\sec\theta=2\cos\theta\) in \([0,2\pi)\).