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Geometry and Trigonometry - Solving Trigonometric Equations

Grade 11IB

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

🔑Concepts

The Unit Circle: A circle with a radius of 11 centered at the origin (0,0)(0,0) where the coordinates of any point PP are (cosθ,sinθ)(\cos \theta, \sin \theta). Visually, as a point moves counter-clockwise from the positive x-axis, the y-coordinate represents the sine value and the x-coordinate represents the cosine value.

The CAST Diagram: A visual tool to identify where trigonometric ratios are positive. In the first quadrant (0 to 9090^\circ or π2\frac{\pi}{2}), All (A) are positive; in the second, Sine (S) is positive; in the third, Tangent (T) is positive; and in the fourth, Cosine (C) is positive.

Reference Angles: For any angle θ\theta, the reference angle θref\theta_{ref} is the acute angle formed with the x-axis. To find solutions, we use θref\theta_{ref} to determine values in other quadrants: Quadrant II is πθref\pi - \theta_{ref}, Quadrant III is π+θref\pi + \theta_{ref}, and Quadrant IV is 2πθref2\pi - \theta_{ref}.

Periodicity and Domain: Trigonometric functions repeat their values over intervals (2π2\pi for sin\sin and cos\cos, π\pi for tan\tan). When solving equations, always check the specified domain (e.g., 0x2π0 \le x \le 2\pi) to ensure all possible solutions are found.

Solving by Substitution: For equations like sin(2x)=k\sin(2x) = k, we substitute u=2xu = 2x and adjust the domain (if x[0,π]x \in [0, \pi], then u[0,2π]u \in [0, 2\pi]). This visualizes a 'horizontal shrink' where the graph completes cycles more quickly, resulting in more solutions within the original range.

Quadratic Trigonometric Equations: Equations in the form asin2x+bsinx+c=0a\sin^2 x + b\sin x + c = 0 can be solved by treating sinx\sin x as a variable (e.g., let u=sinxu = \sin x). Visually, this is like finding the x-intercepts of a parabola, but checking if the resulting uu values fall within the valid range of [1,1][-1, 1].

📐Formulae

sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1

tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}

sin(πθ)=sinθ\sin(\pi - \theta) = \sin \theta

cos(2πθ)=cosθ\cos(2\pi - \theta) = \cos \theta

tan(π+θ)=tanθ\tan(\pi + \theta) = \tan \theta

General solution for sinx=k:x=nπ+(1)narcsink\text{General solution for } \sin x = k: x = n\pi + (-1)^n \arcsin k

General solution for cosx=k:x=2nπ±arccosk\text{General solution for } \cos x = k: x = 2n\pi \pm \arccos k

💡Examples

Problem 1:

Solve 2sin(x)3=02\sin(x) - \sqrt{3} = 0 for 0x2π0 \le x \le 2\pi.

Solution:

Step 1: Isolate sinx\sin x. 2sinx=3    sinx=322\sin x = \sqrt{3} \implies \sin x = \frac{\sqrt{3}}{2} Step 2: Find the reference angle. Since sin(π3)=32\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}, the reference angle is xref=π3x_{ref} = \frac{\pi}{3}. Step 3: Identify quadrants where sine is positive (Quadrants I and II). Quadrant I: x=π3x = \frac{\pi}{3} Quadrant II: x=ππ3=2π3x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} Step 4: Verify the domain. Both π3\frac{\pi}{3} and 2π3\frac{2\pi}{3} are within [0,2π][0, 2\pi]. Final Answer: x=π3,2π3x = \frac{\pi}{3}, \frac{2\pi}{3}

Explanation:

Isolate the trigonometric term and use the unit circle to find the angles in the given domain where the sine value is positive.

Problem 2:

Solve 2cos2x+cosx1=02\cos^2 x + \cos x - 1 = 0 for 0x3600^\circ \le x \le 360^\circ.

Solution:

Step 1: Let u=cosxu = \cos x. The equation becomes 2u2+u1=02u^2 + u - 1 = 0. Step 2: Factor the quadratic. (2u1)(u+1)=0(2u - 1)(u + 1) = 0 So, u=12u = \frac{1}{2} or u=1u = -1. Step 3: Solve for xx. Case 1: cosx=12\cos x = \frac{1}{2}. The reference angle is 6060^\circ. Cosine is positive in Q1 and Q4. x=60x = 60^\circ and x=36060=300x = 360^\circ - 60^\circ = 300^\circ. Case 2: cosx=1\cos x = -1. On the unit circle, this occurs at 180180^\circ. Final Answer: x=60,180,300x = 60^\circ, 180^\circ, 300^\circ

Explanation:

This is a quadratic equation in terms of cosx\cos x. Factor the quadratic expression and solve for xx for each factor.