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JEE Mains Maths · Trigonometric Equations

Range and Existence of Solutions

An equation f(x) = k has a solution exactly when k lies in the range of f, so finding the range answers the question.

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Why this matters

Ten PYQs, eight of them multiple choice, and three from 2026. Six find the range of one side, to fix the values of a constant for which a solution exists or to show there is none; four write a cos x + b sin x as one sine or cosine and solve it. Two ideas cover the page.

Concept 1 of 2: The range decides whether a solution exists

To find the values of kk for which f(x)=kf(x)=k has a solution, find the range of ff. Write ff in one ratio, say t=cos⁡x∈[−1,1]t=\cos x\in[-1,1], and find the range of the polynomial in tt on [−1,1][-1,1]; watch where the vertex falls. The same test proves that an equation has no solution: if one side can never reach the other, the count is 0.

Definition

  • f(x)=kf(x)=k has a solution exactly when kk lies in the range of ff.
  • For g(t)=at2+bt+cg(t)=at^2+bt+c on [−1,1][-1,1]: compare g(−1)g(-1), g(1)g(1) and, if −b2a-\frac{b}{2a} lies inside, the vertex value.
  • A constant inside the equation: solve for it in terms of tt first.
  • sin⁡x\sin x or cos⁡x\cos x outside [−1,1][-1,1], or esin⁡xe^{\sin x} outside [e−1,e][e^{-1},e]: no solution.

Existence

f(x)=k has a solution exactly when min⁡f≤k≤max⁡ff(x)=k\ \text{has a solution exactly when}\ \min f\le k\le\max f

Worked example

For which kk has cos⁡2x+2sin⁡x=k\cos2x+2\sin x=k a real solution?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q62Moderate

Example 1 · Trigonometric Equations · Range and Existence of Solutions

The sum of all the integral values of p such that the equation 3sin⁡2x+12cos⁡x−3=p,x∈R3\sin^{2}x+ 12\cos x- 3 =p,x\in R, has at least one solution, is :

The vertex may be outside

The range of at2+bt+cat^2+bt+c on [−1,1][-1,1] does not always reach the vertex value. If the vertex lies outside [−1,1][-1,1], the extremes are g(−1)g(-1) and g(1)g(1).
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Concept 2 of 2: a cos x + b sin x = c by the auxiliary angle

Divide by R=a2+b2R=\sqrt{a^2+b^2}. The left side becomes Rcos⁡(x−φ)R\cos(x-\varphi), one ratio of one angle. Then solve cos⁡(x−φ)=cR\cos(x-\varphi)=\frac cR: there is no solution if ∣c∣>R|c|>R, and otherwise two per period. When φ\varphi is not a standard angle, use t=tan⁡x2t=\tan\frac x2 instead and solve a quadratic in tt.

Definition

  • acos⁡x+bsin⁡x=Rcos⁡(x−φ)a\cos x+b\sin x=R\cos(x-\varphi), with R=a2+b2R=\sqrt{a^2+b^2} and tan⁡φ=ba\tan\varphi=\frac ba.
  • A solution exists exactly when ∣c∣≤R|c|\le R.
  • sin⁡x=2t1+t2\sin x=\frac{2t}{1+t^2}, cos⁡x=1−t21+t2\cos x=\frac{1-t^2}{1+t^2} with t=tan⁡x2t=\tan\frac x2.
  • The tt substitution misses x=πx=\pi; test it separately.

Auxiliary angle

acos⁡x+bsin⁡x=a2+b2 cos⁡(x−φ),tan⁡φ=baa\cos x+b\sin x=\sqrt{a^2+b^2}\,\cos(x-\varphi),\qquad\tan\varphi=\frac ba

Worked example

Find all x∈[0,2π]x\in[0,2\pi] with sin⁡x+cos⁡x=1\sin x+\cos x=1.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q64Moderate

Example 2 · Trigonometric Equations · Range and Existence of Solutions

Let S={θ∈(−2π,2π):cos⁡θ+1=3sin⁡θ}S = \{\theta\in ( - 2\pi,2\pi):\cos\theta+ 1 =\sqrt{3}\sin\theta\}. Then ∑θ∈Sθ\sum_{\theta\in S} \theta is equal to:

Squaring adds roots

Squaring acos⁡x=c−bsin⁡xa\cos x=c-b\sin x to get a quadratic in sin⁡x\sin x also brings in the roots of acos⁡x=−(c−bsin⁡x)a\cos x=-(c-b\sin x). Check each root in the original equation, or use the auxiliary angle, which needs no check.
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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)

  • The range decides whether a solution exists

    Existence

    f(x)=k has a solution exactly when min⁡f≤k≤max⁡ff(x)=k\ \text{has a solution exactly when}\ \min f\le k\le\max f
  • a cos x + b sin x = c by the auxiliary angle

    Auxiliary angle

    acos⁡x+bsin⁡x=a2+b2 cos⁡(x−φ),tan⁡φ=baa\cos x+b\sin x=\sqrt{a^2+b^2}\,\cos(x-\varphi),\qquad\tan\varphi=\frac ba

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.

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