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Trigonometric Functions - Trigonometric Equations

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Trigonometric equations are equations involving one or more trigonometric functions of unknown angles, such as sin⁡x=12\sin x = \frac{1}{2} or cos⁡2x−sin⁡x=1\cos^2 x - \sin x = 1.

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The principal solutions of a trigonometric equation are the solutions where the angle xx lies in the interval 0≤x<2π0 \le x < 2\pi. Visually, these represent the specific points within one full rotation of the unit circle where the function meets a given value.

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The general solution of a trigonometric equation is an expression involving an integer nn that represents all possible solutions. Because trigonometric functions are periodic, their graphs repeat at regular intervals (every 2π2\pi for sine and cosine, and every π\pi for tangent), leading to an infinite number of solutions.

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For the equation sin⁡x=0\sin x = 0, the solutions occur where the sine graph crosses the x-axis, which happens at every multiple of π\pi. Visually, these are the points on the unit circle where the y-coordinate is zero, specifically at 0,π,2π,...0, \pi, 2\pi, ... etc.

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For the equation cos⁡x=0\cos x = 0, the solutions occur at odd multiples of π2\frac{\pi}{2}. On a graph, these are the x-intercepts of the cosine wave, and on a unit circle, they represent the points where the x-coordinate is zero (the top and bottom of the circle).

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The CAST rule or Quadrant system helps determine the signs of trigonometric functions across the four quadrants (00 to 2π2\pi). This is vital for finding principal solutions; for example, if tan⁡x\tan x is positive, the solutions must lie in the first quadrant (where all are positive) or the third quadrant (where only tangent and cotangent are positive).

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General solutions for sin⁡x=sin⁡y\sin x = \sin y involve the term (−1)n(-1)^n. This alternating sign accounts for the fact that sine is positive in the first and second quadrants, creating a 'zigzag' pattern of solutions across the periodic wave.

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General solutions for cos⁡x=cos⁡y\cos x = \cos y use the ±\pm notation because the cosine function is symmetric about the y-axis (even function), meaning cos⁡(x)=cos⁡(−x)\cos(x) = \cos(-x). This visual symmetry results in two solution paths for every 2π2\pi rotation.

📐Formulae

sin⁡x=0  ⟹  x=nπ, where n∈Z\sin x = 0 \implies x = n\pi, \text{ where } n \in \mathbb{Z}

cos⁡x=0  ⟹  x=(2n+1)π2, where n∈Z\cos x = 0 \implies x = (2n + 1)\frac{\pi}{2}, \text{ where } n \in \mathbb{Z}

tan⁡x=0  ⟹  x=nπ, where n∈Z\tan x = 0 \implies x = n\pi, \text{ where } n \in \mathbb{Z}

sin⁡x=sin⁡α  ⟹  x=nπ+(−1)nα, where n∈Z\sin x = \sin \alpha \implies x = n\pi + (-1)^n \alpha, \text{ where } n \in \mathbb{Z}

cos⁡x=cos⁡α  ⟹  x=2nπ±α, where n∈Z\cos x = \cos \alpha \implies x = 2n\pi \pm \alpha, \text{ where } n \in \mathbb{Z}

tan⁡x=tan⁡α  ⟹  x=nπ+α, where n∈Z\tan x = \tan \alpha \implies x = n\pi + \alpha, \text{ where } n \in \mathbb{Z}

💡Examples

Problem 1:

Find the principal and general solutions of the equation sin⁡x=32\sin x = \frac{\sqrt{3}}{2}.

Solution:

Step 1: We know that sin⁡π3=32\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}. Since 32\frac{\sqrt{3}}{2} is positive, sin⁡x\sin x is positive in the I and II quadrants. Step 2: In Quadrant I, x=π3x = \frac{\pi}{3}. Step 3: In Quadrant II, x=π−π3=2π3x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}. Step 4: Therefore, the principal solutions are x=π3x = \frac{\pi}{3} and x=2π3x = \frac{2\pi}{3}. Step 5: For the general solution, use the formula x=nπ+(−1)nαx = n\pi + (-1)^n \alpha with α=π3\alpha = \frac{\pi}{3}. Result: x=nπ+(−1)nπ3,n∈Zx = n\pi + (-1)^n \frac{\pi}{3}, n \in \mathbb{Z}.

Explanation:

Identify the base angle α\alpha in the first quadrant, then use the quadrant rules to find principal solutions within [0,2π)[0, 2\pi), and finally apply the general formula for sine.

Problem 2:

Solve cos⁡2x=cos⁡x\cos 2x = \cos x.

Solution:

Step 1: Use the general solution formula for cos⁡θ=cos⁡α\cos \theta = \cos \alpha, which is θ=2nπ±α\theta = 2n\pi \pm \alpha. Step 2: Here, θ=2x\theta = 2x and α=x\alpha = x. So, 2x=2nπ±x2x = 2n\pi \pm x. Step 3: Case 1: 2x=2nπ+x  ⟹  x=2nπ2x = 2n\pi + x \implies x = 2n\pi. Step 4: Case 2: 2x=2nπ−x  ⟹  3x=2nπ  ⟹  x=2nπ32x = 2n\pi - x \implies 3x = 2n\pi \implies x = \frac{2n\pi}{3}. Step 5: Combining these, the general solution is x=2nπx = 2n\pi or x=2nπ3,n∈Zx = \frac{2n\pi}{3}, n \in \mathbb{Z}.

Explanation:

Instead of converting to a quadratic, applying the general solution formula directly is more efficient. We split the equation into two cases based on the plus-minus sign.