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JEE Mains Maths · Trigonometric Equations

Quadratic in One Ratio

Equations that become a quadratic in sin θ, cos θ, sec θ or csc θ: solve it, discard the root outside the range, then count the angles in the interval.

All Trigonometric Equations notesJEE Mains Maths strategy

Why this matters

Sixteen PYQs, ten of them multiple choice, and one from 2026. Ten reduce to a quadratic in one ratio and count or add its roots in an interval; six add a condition — two equations at once, a domain limit, or an interval sized to hold an exact number of roots. Two ideas cover the page.

Concept 1 of 2: Solve for the ratio, then count per period

Replace cos⁡2θ\cos2\theta, sin⁡2θ\sin^2\theta or tan⁡2θ\tan^2\theta so that only one ratio is left. The equation becomes a quadratic in that ratio. A root outside [−1,1][-1,1] for sin⁡\sin or cos⁡\cos, or inside (−1,1)(-1,1) for sec⁡\sec or csc⁡\csc, gives no angle. Each value that survives is taken twice in every period of 2π2\pi, so count by periods and test the endpoints separately.

Definition

  • cos⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta; tan⁡2θ=sec⁡2θ−1\tan^2\theta=\sec^2\theta-1.
  • Reject ∣sin⁡θ∣>1|\sin\theta|>1, ∣cos⁡θ∣>1|\cos\theta|>1, ∣sec⁡θ∣<1|\sec\theta|<1, ∣csc⁡θ∣<1|\csc\theta|<1.
  • cos⁡θ=c\cos\theta=c with ∣c∣<1|c|<1: two solutions in each full period of length 2π2\pi.
  • Substitute each endpoint of the interval to see if it is a solution.

General solutions

sin⁡θ=sin⁡α⇒θ=nπ+(−1)nα;cos⁡θ=cos⁡α⇒θ=2nπ±α\sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha;\qquad \cos\theta=\cos\alpha\Rightarrow\theta=2n\pi\pm\alpha

Worked example

How many solutions has 2sin⁡2θ+3cos⁡θ=02\sin^2\theta+3\cos\theta=0 in [0,4π][0,4\pi]?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q66Moderate

Example 1 · Trigonometric Equations · Quadratic in One Ratio

Number of solutions of 3cos⁡2θ+8cos⁡θ+33=0,θ∈[−3π,2π]\sqrt{3}\cos2\theta + 8\cos\theta + 3\sqrt{3}= 0,\theta \in \lbrack - 3\pi,2\pi\rbrack is:

Values ±1 and the endpoints

cos⁡θ=±1\cos\theta=\pm1 or sin⁡θ=±1\sin\theta=\pm1 is reached once per period, not twice. And a closed interval can hold a solution at each end: cos⁡θ=−1\cos\theta=-1 on [−π,π][-\pi,\pi] holds at both −π-\pi and π\pi.
Drill 9 more on solve for the ratio, then count per period

Concept 2 of 2: Two equations, domain limits and a sized interval

Some questions add a condition to the quadratic. When two equations must both hold, solve each for the ratio and keep the common value. When tan⁡\tan, sec⁡\sec or a logarithm appears, drop any root where it is undefined. When the interval is unknown and must hold exactly kk roots, list the roots in order from the left end and stop at the kk-th.

Definition

  • Two equations: find the ratio values of each, and keep those in both.
  • Remove roots with cos⁡θ=0\cos\theta=0 if tan⁡θ\tan\theta or sec⁡θ\sec\theta appears. A log base must be positive and not 1.
  • Exactly kk roots in [0,L][0,L]: LL is at least the kk-th root and less than the (k+1)(k+1)-th.
  • Do not cancel a ratio; take it out as a factor.

Roots of cos θ = c in order from 0

α, 2π−α, 2π+α, 4π−α, …(α=cos⁡−1c∈(0,π))\alpha,\ 2\pi-\alpha,\ 2\pi+\alpha,\ 4\pi-\alpha,\ \dots\qquad(\alpha=\cos^{-1}c\in(0,\pi))

Worked example

Find the sum of all θ∈[0,2π]\theta\in[0,2\pi] satisfying both 2cos⁡2θ−cos⁡θ−1=02\cos^2\theta-\cos\theta-1=0 and cos⁡2θ=−12\cos2\theta=-\frac12.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q143Moderate

Example 2 · Trigonometric Equations · Quadratic in One Ratio

The sum of all values of θ∈[0,2π]\theta \in \lbrack 0,2\pi\rbrack satisfying 2sin⁡2θ=cos⁡2θ2\sin^{2}\theta = \cos2\theta and 2cos⁡2θ=3sin⁡θ2\cos^{2}\theta = 3\sin\theta is

Do not cancel a ratio

In sin⁡θcos⁡θ=sin⁡θ\sin\theta\cos\theta=\sin\theta, cancelling sin⁡θ\sin\theta leaves cos⁡θ=1\cos\theta=1 and loses θ=π\theta=\pi. Write sin⁡θ(cos⁡θ−1)=0\sin\theta(\cos\theta-1)=0 and keep both factors.
Drill 5 more on two equations, domain limits and a sized interval

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Solve for the ratio, then count per period

    General solutions

    sin⁡θ=sin⁡α⇒θ=nπ+(−1)nα;cos⁡θ=cos⁡α⇒θ=2nπ±α\sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha;\qquad \cos\theta=\cos\alpha\Rightarrow\theta=2n\pi\pm\alpha
  • Two equations, domain limits and a sized interval

    Roots of cos θ = c in order from 0

    α, 2π−α, 2π+α, 4π−α, …(α=cos⁡−1c∈(0,π))\alpha,\ 2\pi-\alpha,\ 2\pi+\alpha,\ 4\pi-\alpha,\ \dots\qquad(\alpha=\cos^{-1}c\in(0,\pi))

Watch out for (2)

Test yourself on Trigonometric Equations

15 past JEE Mains questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Trigonometric Equations chapter testDrill all 16 questions