JEE Challenger
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MathMedAdv 2010 • Paper 1

Simultaneous Solutions of Trigonometric Equations in Specified Interval

The number of values of θ\theta in the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) such that θ≠nπ5\theta \neq \frac{n\pi}{5} for n=0,±1,±2n = 0, \pm 1, \pm 2 and tan⁡θ=cot⁡5θ\tan\theta = \cot 5\theta as well as sin⁡2θ=cos⁡4θ\sin 2\theta = \cos 4\theta is

Official Numerical Answer3

Topics & Concepts

Trigonometric FunctionsTrigonometric EquationsTrigonometric Ratios and Identities

Step-by-Step Solution

To find the number of values of θ∈(−π2,π2)\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) satisfying the given conditions, we solve the system of trigonometric equations step-by-step:

Step 1: Solve the first equation tan⁡θ=cot⁡5θ\tan\theta = \cot 5\theta

The equation is defined for:
θ≠π2+kπand5θ≠nπ  ⟹  θ≠nπ5\theta \neq \frac{\pi}{2} + k\pi \quad \text{and} \quad 5\theta \neq n\pi \implies \theta \neq \frac{n\pi}{5}
Using the identity cot⁡(A)=tan⁡(π2−A)\cot(A) = \tan\left(\frac{\pi}{2} - A\right), we have:
tan⁡θ=tan⁡(π2−5θ)\tan\theta = \tan\left(\frac{\pi}{2} - 5\theta\right)
The general solution is:
θ=kπ+(π2−5θ),k∈Z\theta = k\pi + \left(\frac{\pi}{2} - 5\theta\right), \quad k \in \mathbb{Z}
6θ=kπ+π2=(2k+1)π26\theta = k\pi + \frac{\pi}{2} = (2k + 1)\frac{\pi}{2}
θ=(2k+1)π12,k∈Z\theta = \frac{(2k + 1)\pi}{12}, \quad k \in \mathbb{Z}

Since θ∈(−π2,π2)\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), we find the allowed values of 2k+12k + 1:
−π2<(2k+1)π12<π2  ⟹  −6<2k+1<6-\frac{\pi}{2} < \frac{(2k + 1)\pi}{12} < \frac{\pi}{2} \implies -6 < 2k + 1 < 6
The possible odd integers for 2k+12k + 1 are:
2k+1∈{−5,−3,−1,1,3,5}2k + 1 \in \{-5, -3, -1, 1, 3, 5\}
Thus, the candidate values for θ\theta are:
θ∈{−5π12,−π4,−π12,π12,π4,5π12}\theta \in \left\{ -\frac{5\pi}{12}, -\frac{\pi}{4}, -\frac{\pi}{12}, \frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12} \right\}

Step 2: Solve the second equation sin⁡2θ=cos⁡4θ\sin 2\theta = \cos 4\theta

Using the identity sin⁡(A)=cos⁡(π2−A)\sin(A) = \cos\left(\frac{\pi}{2} - A\right), we can write:
cos⁡4θ=cos⁡(π2−2θ)\cos 4\theta = \cos\left(\frac{\pi}{2} - 2\theta\right)
The general solution for cos⁡X=cos⁡Y\cos X = \cos Y is X=2mπ±YX = 2m\pi \pm Y, where m∈Zm \in \mathbb{Z}:

  1. Case 1:
    4θ=2mπ+(π2−2θ)4\theta = 2m\pi + \left(\frac{\pi}{2} - 2\theta\right)
    6θ=2mπ+π2=(4m+1)π26\theta = 2m\pi + \frac{\pi}{2} = (4m + 1)\frac{\pi}{2}
    θ=(4m+1)π12\theta = \frac{(4m + 1)\pi}{12}
    Comparing this with θ=(2k+1)π12\theta = \frac{(2k + 1)\pi}{12}, we must have:
    2k+1≡1(mod4)2k + 1 \equiv 1 \pmod 4
    Among the candidates {−5,−3,−1,1,3,5}\{-5, -3, -1, 1, 3, 5\}:

    • For 2k+1=−32k+1 = -3: −3=4(−1)+1  ⟹  θ=−3π12=−π4-3 = 4(-1) + 1 \implies \theta = -\frac{3\pi}{12} = -\frac{\pi}{4}
    • For 2k+1=12k+1 = 1: 1=4(0)+1  ⟹  θ=π121 = 4(0) + 1 \implies \theta = \frac{\pi}{12}
    • For 2k+1=52k+1 = 5: 5=4(1)+1  ⟹  θ=5π125 = 4(1) + 1 \implies \theta = \frac{5\pi}{12}

  2. Case 2:
    4θ=2mπ−(π2−2θ)4\theta = 2m\pi - \left(\frac{\pi}{2} - 2\theta\right)
    2θ=2mπ−π2=(4m−1)π22\theta = 2m\pi - \frac{\pi}{2} = (4m - 1)\frac{\pi}{2}
    θ=(4m−1)π4\theta = \frac{(4m - 1)\pi}{4}
    For θ∈(−π2,π2)\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right):
    −12<4m−14<12  ⟹  −1<4m<3  ⟹  m=0-\frac{1}{2} < \frac{4m - 1}{4} < \frac{1}{2} \implies -1 < 4m < 3 \implies m = 0
    This yields θ=−π4\theta = -\frac{\pi}{4}, which is already included in Case 1.

Step 3: Check domain restrictions

The condition states that θ≠nπ5\theta \neq \frac{n\pi}{5} for n=0,±1,±2n = 0, \pm 1, \pm 2:

  1. θ=−π4≠nπ5\theta = -\frac{\pi}{4} \neq \frac{n\pi}{5} since −14≠n5  ⟹  n=−54∉Z-\frac{1}{4} \neq \frac{n}{5} \implies n = -\frac{5}{4} \notin \mathbb{Z}
  2. θ=π12≠nπ5\theta = \frac{\pi}{12} \neq \frac{n\pi}{5} since 112≠n5  ⟹  n=512∉Z\frac{1}{12} \neq \frac{n}{5} \implies n = \frac{5}{12} \notin \mathbb{Z}
  3. θ=5π12≠nπ5\theta = \frac{5\pi}{12} \neq \frac{n\pi}{5} since 512≠n5  ⟹  n=2512∉Z\frac{5}{12} \neq \frac{n}{5} \implies n = \frac{25}{12} \notin \mathbb{Z}

All three values satisfy all the conditions of the problem.

Thus, the simultaneous solutions are:
θ∈{−π4,π12,5π12}\theta \in \left\{ -\frac{\pi}{4}, \frac{\pi}{12}, \frac{5\pi}{12} \right\}

The total number of values of θ\theta is 3.

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