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Trigonometry - Trigonometric Equations (General Solutions)

Grade 11ICSE

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

🔑Concepts

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Trigonometric Equations are equations involving trigonometric functions of unknown angles, such as sin⁡x=12\sin x = \frac{1}{2}. Unlike algebraic equations, these often have infinitely many solutions because trigonometric functions are periodic, meaning their values repeat at regular intervals.

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The Principal Solution of a trigonometric equation refers to the values of the unknown angle that lie within the interval 0≤θ<2π0 \le \theta < 2\pi. Visually, these are the specific points on a unit circle or within the first cycle of a wave graph that satisfy the equation.

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The General Solution is a mathematical expression that represents all possible solutions to a trigonometric equation. It incorporates an integer nn (n∈Zn \in \mathbb{Z}) to account for the infinite repetitions caused by the periodicity of the function.

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Periodicity and Quadrants: The sign of a trigonometric function depends on the quadrant in which the angle lies (ASTC rule: All-Positive in Q1, Sine in Q2, Tangent in Q3, Cosine in Q4). When solving sin⁡θ=k\sin \theta = k, the general solution θ=nπ+(−1)nα\theta = n\pi + (-1)^n \alpha accounts for the symmetry of the sine wave across the vertical axis in the Cartesian plane.

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General Solution for Sine: For the equation sin⁡θ=sin⁡α\sin \theta = \sin \alpha, the general solution is θ=nπ+(−1)nα\theta = n\pi + (-1)^n \alpha. This formula accounts for the fact that sine is positive in both the first and second quadrants and repeats every 2π2\pi radians.

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General Solution for Cosine: For the equation cos⁡θ=cos⁡α\cos \theta = \cos \alpha, the general solution is θ=2nπ±α\theta = 2n\pi \pm \alpha. This reflects the symmetry of the cosine graph across the x-axis, where values are identical for positive and negative angles in the first and fourth quadrants.

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General Solution for Tangent: For the equation tan⁡θ=tan⁡α\tan \theta = \tan \alpha, the general solution is θ=nπ+α\theta = n\pi + \alpha. Since the tangent function has a period of π\pi, the values repeat every half-circle, appearing as a series of parallel curves separated by vertical asymptotes.

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Squared Trigonometric Equations: Equations of the form sin⁡2θ=sin⁡2α\sin^2 \theta = \sin^2 \alpha, cos⁡2θ=cos⁡2α\cos^2 \theta = \cos^2 \alpha, or tan⁡2θ=tan⁡2α\tan^2 \theta = \tan^2 \alpha all share the same general solution: θ=nπ±α\theta = n\pi \pm \alpha. Visually, this covers four points on the unit circle, one in each quadrant, placed symmetrically relative to the axes.

📐Formulae

If sin⁡θ=0\sin \theta = 0, then θ=nπ,n∈Z\theta = n\pi, n \in \mathbb{Z}

If cos⁡θ=0\cos \theta = 0, then θ=(2n+1)π2,n∈Z\theta = (2n + 1)\frac{\pi}{2}, n \in \mathbb{Z}

If tan⁡θ=0\tan \theta = 0, then θ=nπ,n∈Z\theta = n\pi, n \in \mathbb{Z}

If sin⁡θ=sin⁡α\sin \theta = \sin \alpha, then θ=nπ+(−1)nα,n∈Z\theta = n\pi + (-1)^n \alpha, n \in \mathbb{Z}

If cos⁡θ=cos⁡α\cos \theta = \cos \alpha, then θ=2nπ±α,n∈Z\theta = 2n\pi \pm \alpha, n \in \mathbb{Z}

If tan⁡θ=tan⁡α\tan \theta = \tan \alpha, then θ=nπ+α,n∈Z\theta = n\pi + \alpha, n \in \mathbb{Z}

If sin⁡2θ=sin⁡2α\sin^2 \theta = \sin^2 \alpha, then θ=nπ±α,n∈Z\theta = n\pi \pm \alpha, n \in \mathbb{Z}

If cos⁡2θ=cos⁡2α\cos^2 \theta = \cos^2 \alpha, then θ=nπ±α,n∈Z\theta = n\pi \pm \alpha, n \in \mathbb{Z}

If tan⁡2θ=tan⁡2α\tan^2 \theta = \tan^2 \alpha, then θ=nπ±α,n∈Z\theta = n\pi \pm \alpha, n \in \mathbb{Z}

💡Examples

Problem 1:

Find the general solution of the equation 3sec⁡2θ=2\sqrt{3} \sec 2\theta = 2.

Solution:

Step 1: Convert the equation to a basic trigonometric form. sec⁡2θ=23\sec 2\theta = \frac{2}{\sqrt{3}} cos⁡2θ=32\cos 2\theta = \frac{\sqrt{3}}{2}

Step 2: Find the principal value α\alpha. We know cos⁡π6=32\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, so let α=π6\alpha = \frac{\pi}{6}.

Step 3: Apply the general solution formula for cosine. 2θ=2nπ±π62\theta = 2n\pi \pm \frac{\pi}{6}

Step 4: Solve for θ\theta. θ=nπ±π12,n∈Z\theta = n\pi \pm \frac{\pi}{12}, n \in \mathbb{Z}

Explanation:

We first express the secant function in terms of cosine. Then, we identify the basic angle α\alpha for which the cosine value is 32\frac{\sqrt{3}}{2}. Finally, we use the general solution formula for cos⁡θ=cos⁡α\cos \theta = \cos \alpha and divide by the coefficient of θ\theta.

Problem 2:

Solve for the general solution: 2sin⁡2x+sin⁡x−1=02\sin^2 x + \sin x - 1 = 0.

Solution:

Step 1: Treat the equation as a quadratic in terms of sin⁡x\sin x. Let y=sin⁡xy = \sin x, then 2y2+y−1=02y^2 + y - 1 = 0.

Step 2: Factorize the quadratic equation. 2y2+2y−y−1=02y^2 + 2y - y - 1 = 0 2y(y+1)−1(y+1)=02y(y + 1) - 1(y + 1) = 0 (2y−1)(y+1)=0(2y - 1)(y + 1) = 0

Step 3: Solve for sin⁡x\sin x. Case 1: sin⁡x=12  ⟹  sin⁡x=sin⁡π6\sin x = \frac{1}{2} \implies \sin x = \sin \frac{\pi}{6} x=nπ+(−1)nπ6x = n\pi + (-1)^n \frac{\pi}{6} Case 2: sin⁡x=−1  ⟹  sin⁡x=sin⁡(−π2)\sin x = -1 \implies \sin x = \sin (-\frac{\pi}{2}) x=kπ+(−1)k(−π2)x = k\pi + (-1)^k (-\frac{\pi}{2})

Step 4: Combine the solutions. x=nπ+(−1)nπ6x = n\pi + (-1)^n \frac{\pi}{6} or x=kπ−(−1)kπ2x = k\pi - (-1)^k \frac{\pi}{2} where n,k∈Zn, k \in \mathbb{Z}.

Explanation:

This is a quadratic trigonometric equation. We factor it like a standard polynomial to find two possible values for sin⁡x\sin x. Each value leads to a separate general solution using the sine formula.