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Trigonometric equations

A trigonometric equation is an equation involving trigonometric functions of an unknown angle. Because these functions are periodic, every such equation has infinitely many solutions, and we must learn to write all of them in one expression , the general solution.

Principal value vs general solution

If sin⁡x=12\sin x = \dfrac{1}{2}, then x=π6x = \dfrac{\pi}{6} works. But so does 5π6\dfrac{5\pi}{6} (the same height on the other side of the unit circle), and also π6+2π\dfrac{\pi}{6} + 2\pi, 5π6+2π\dfrac{5\pi}{6} + 2\pi, π6−2π\dfrac{\pi}{6} - 2\pi, and so on. The general solution captures every one of these.

The three master rules

For each basic equation we write the general solution in closed form.

sin⁡x=sin⁡α\sin x = \sin\alpha

x=nπ+(−1)nα,n∈Z.x = n\pi + (-1)^n \alpha, \quad n \in \mathbb{Z}.

Verify at n=0n = 0: x=αx = \alpha. At n=1n = 1: x=π−αx = \pi - \alpha. At n=2n = 2: x=2π+αx = 2\pi + \alpha. Etc.

cos⁡x=cos⁡α\cos x = \cos\alpha

x=2nπ±α,n∈Z.x = 2n\pi \pm \alpha, \quad n \in \mathbb{Z}.

tan⁡x=tan⁡α\tan x = \tan\alpha

x=nπ+α,n∈Z.x = n\pi + \alpha, \quad n \in \mathbb{Z}.

(Period of tan⁡\tan is π\pi, not 2π2\pi.)

Three special cases

EquationGeneral solution
sin⁡x=0\sin x = 0x=nπx = n\pi
cos⁡x=0\cos x = 0x=(2n+1)π2x = (2n+1)\dfrac{\pi}{2}
tan⁡x=0\tan x = 0x=nπx = n\pi
sin⁡x=1\sin x = 1x=(4n+1)π2x = (4n + 1)\dfrac{\pi}{2}
sin⁡x=−1\sin x = -1x=(4n−1)π2x = (4n - 1)\dfrac{\pi}{2}
cos⁡x=1\cos x = 1x=2nπx = 2n\pi
cos⁡x=−1\cos x = -1x=(2n+1)πx = (2n + 1)\pi

Strategy for general equations

  1. Reduce to a basic form. Use identities to get sin⁡=sin⁡\sin = \sin, cos⁡=cos⁡\cos = \cos, or tan⁡=tan⁡\tan = \tan.
  2. Square only when necessary. Squaring can introduce extraneous solutions; check at the end.
  3. Factor. Quadratics in sin⁡x\sin x or cos⁡x\cos x factor and split into two equations.
  4. Write the general solution. Apply the master rules and combine.

Solving asin⁡x+bcos⁡x=ca \sin x + b \cos x = c

Convert to Rsin⁡(x+ϕ)=cR\sin(x + \phi) = c with R=a2+b2R = \sqrt{a^2 + b^2}. The equation has solutions iff ∣c∣≤R|c| \le R. If yes, sin⁡(x+ϕ)=c/R\sin(x + \phi) = c/R, and we solve as sin⁡α=c/R\sin\alpha = c/R.

Worked examples

Example 1. Solve sin⁡x=32\sin x = \dfrac{\sqrt{3}}{2}.

sin⁡x=sin⁡π3\sin x = \sin\dfrac{\pi}{3}, so x=nπ+(−1)nπ3x = n\pi + (-1)^n \dfrac{\pi}{3}, n∈Zn \in \mathbb{Z}.

Example 2. Solve 2cos⁡2x−1=02\cos^2 x - 1 = 0.

cos⁡2x=12\cos^2 x = \dfrac{1}{2}, so cos⁡x=±12\cos x = \pm \dfrac{1}{\sqrt{2}}. The two cases:

  • cos⁡x=12⇒x=2nπ±π4\cos x = \dfrac{1}{\sqrt{2}} \Rightarrow x = 2n\pi \pm \dfrac{\pi}{4}.
  • cos⁡x=−12⇒x=2nπ±3π4\cos x = -\dfrac{1}{\sqrt{2}} \Rightarrow x = 2n\pi \pm \dfrac{3\pi}{4}.

Compactly: x=nπ±π4x = n\pi \pm \dfrac{\pi}{4}, n∈Zn \in \mathbb{Z}. (Or use cos⁡2x=0\cos 2x = 0: 2x=(2n+1)π/22x = (2n+1)\pi/2, x=(2n+1)π/4x = (2n+1)\pi/4.)

Example 3. Solve tan⁡x=−1\tan x = -1.

tan⁡x=tan⁡ ⁣(−π4)\tan x = \tan\!\left(-\dfrac{\pi}{4}\right), so x=nπ−π4x = n\pi - \dfrac{\pi}{4}.

Example 4. Solve sin⁡x+cos⁡x=1\sin x + \cos x = 1.

Convert: 2sin⁡ ⁣(x+π4)=1\sqrt{2}\sin\!\left(x + \dfrac{\pi}{4}\right) = 1, so sin⁡ ⁣(x+π4)=12=sin⁡π4\sin\!\left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}} = \sin\dfrac{\pi}{4}.

So x+π4=nπ+(−1)nπ4x + \dfrac{\pi}{4} = n\pi + (-1)^n \dfrac{\pi}{4}. Splitting by parity:

  • nn even: x+π4=2kπ+π4x + \dfrac{\pi}{4} = 2k\pi + \dfrac{\pi}{4}, so x=2kπx = 2k\pi.
  • nn odd: x+π4=(2k+1)π−π4x + \dfrac{\pi}{4} = (2k+1)\pi - \dfrac{\pi}{4}, so x=2kπ+π2x = 2k\pi + \dfrac{\pi}{2}.

Combined: x=2kπx = 2k\pi or x=2kπ+π2x = 2k\pi + \dfrac{\pi}{2} for k∈Zk \in \mathbb{Z}.

Example 5 (harder). Solve sin⁡5x=sin⁡3x\sin 5x = \sin 3x.

Bring to one side and use sum-to-product: sin⁡5x−sin⁡3x=2cos⁡4xsin⁡x=0.\sin 5x - \sin 3x = 2 \cos 4x \sin x = 0. So cos⁡4x=0\cos 4x = 0 or sin⁡x=0\sin x = 0.

  • cos⁡4x=0⇒4x=(2n+1)π2⇒x=(2n+1)π8\cos 4x = 0 \Rightarrow 4x = (2n+1)\dfrac{\pi}{2} \Rightarrow x = (2n+1)\dfrac{\pi}{8}.
  • sin⁡x=0⇒x=mπ\sin x = 0 \Rightarrow x = m\pi.

General solution: x=mπx = m\pi or x=(2n+1)π8x = (2n+1)\dfrac{\pi}{8}.

Try it yourself

  1. Solve cos⁡x=12\cos x = \dfrac{1}{2}.
  2. Solve tan⁡x=3\tan x = \sqrt{3}.
  3. Solve sin⁡2x=sin⁡x\sin 2x = \sin x.
  4. Solve cos⁡3x=cos⁡2x\cos 3x = \cos 2x.
  5. Solve 2sin⁡2x−3sin⁡x+1=02\sin^2 x - 3\sin x + 1 = 0.
  6. Solve sin⁡x+sin⁡3x=0\sin x + \sin 3x = 0.
  7. Solve 3sin⁡x+cos⁡x=1\sqrt{3}\sin x + \cos x = 1.
  8. Solve tan⁡x+cot⁡x=2\tan x + \cot x = 2.
  9. Solve sin⁡2x−cos⁡2x=12\sin^2 x - \cos^2 x = \dfrac{1}{2} (use cos⁡2x=cos⁡2−sin⁡2\cos 2x = \cos^2 - \sin^2).
  10. Find all x∈[0,2π]x \in [0, 2\pi] with sin⁡x=cos⁡x\sin x = \cos x.
  11. Solve cos⁡xcos⁡2x=14\cos x \cos 2x = \dfrac{1}{4}. (Use product-to-sum.)
  12. Solve sin⁡x+sin⁡2x+sin⁡3x=0\sin x + \sin 2x + \sin 3x = 0.

Pitfalls / Tricks

  • For sin⁡x=0\sin x = 0, the general solution is x=nπx = n\pi, not x=2nπx = 2n\pi , the period of sin⁡x\sin x is 2π2\pi, but zeros occur every π\pi.
  • When you square both sides, check the final answers , squaring can introduce spurious roots.
  • tan⁡x\tan x has solutions x=nπ+αx = n\pi + \alpha, without the (−1)n(-1)^n factor (which is only for sin⁡\sin).
  • Insight. Almost every trigonometric equation reduces to sin⁡f(x)=sin⁡g(x)\sin f(x) = \sin g(x) or cos⁡f(x)=cos⁡g(x)\cos f(x) = \cos g(x). Master these two patterns and you can solve nearly any problem.

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