Step-by-Step Examples using the Unit Circle and Identities
Solving trigonometric equations requires isolating the trigonometric function, finding the reference angle, and using the unit circle to determine the angles in the correct quadrants. When equations are more complex, trigonometric identities are used to simplify them first. Let's explore examples ranging from basic linear forms to challenging quadratics.
Find all the solutions of the trigonometric equation: \[ \sqrt{3} \sec(\theta) + 2 = 0 \]
Using the reciprocal identity \( \sec(\theta) = \dfrac{1}{\cos(\theta)} \), we can rewrite the equation to isolate cosine:
\[ \cos(\theta) = -\dfrac{\sqrt{3}}{2} \]
Find the reference angle \( \theta_r \) by taking the absolute value: \( \cos(\theta_r) = \dfrac{\sqrt{3}}{2} \):
\[ \theta_r = \dfrac{\pi}{6} \]
Since \( \cos(\theta) \) is negative, the actual angles must lie in Quadrants II or III.
\[ \text{Quadrant II: } \theta_1 = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6} \]
\[ \text{Quadrant III: } \theta_2 = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6} \]
General Solutions: Add \( 2n\pi \) to account for the periodic nature of cosine (where \(n\) is an integer).
\[ \theta = \mathbf{\dfrac{5\pi}{6} + 2n\pi} \quad \text{and} \quad \theta = \mathbf{\dfrac{7\pi}{6} + 2n\pi} \]
Solve the trigonometric equation: \[ 2 \sin(\theta) = -1 \]
Isolate the sine function by dividing both sides by 2:
\[ \sin(\theta) = -\dfrac{1}{2} \]
The reference angle where sine is \( 1/2 \) is \( \theta_r = \dfrac{\pi}{6} \). Because the sine value is negative, the angles must be in Quadrants III and IV.
\[ \text{Quadrant III: } \theta_1 = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6} \]
\[ \text{Quadrant IV: } \theta_2 = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6} \]
General Solutions:
\[ \theta = \mathbf{\dfrac{7\pi}{6} + 2n\pi} \quad \text{and} \quad \theta = \mathbf{\dfrac{11\pi}{6} + 2n\pi} \]
Solve the following equation involving a phase shift and multiple angle: \[ \sqrt{2} \cos\left(3x + \dfrac{\pi}{4}\right) = -1 \]
First, substitute the inside argument with a temporary variable. Let \( \theta = 3x + \dfrac{\pi}{4} \). Isolate cosine:
\[ \cos(\theta) = -\dfrac{1}{\sqrt{2}} \]
The reference angle is \( \theta_r = \dfrac{\pi}{4} \). Cosine is negative in Quadrants II and III. Find the solutions for \(\theta\):
\[ \theta_1 = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4} + 2n\pi \]
\[ \theta_2 = \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4} + 2n\pi \]
Now, substitute \( 3x + \dfrac{\pi}{4} \) back in for \(\theta\) and solve for \(x\).
First Equation:
\[ 3x + \dfrac{\pi}{4} = \dfrac{3\pi}{4} + 2n\pi \]
\[ 3x = \dfrac{2\pi}{4} + 2n\pi = \dfrac{\pi}{2} + 2n\pi \]
\[ x_1 = \mathbf{\dfrac{\pi}{6} + \dfrac{2n\pi}{3}} \]
Second Equation:
\[ 3x + \dfrac{\pi}{4} = \dfrac{5\pi}{4} + 2n\pi \]
\[ 3x = \dfrac{4\pi}{4} + 2n\pi = \pi + 2n\pi \]
\[ x_2 = \mathbf{\dfrac{\pi}{3} + \dfrac{2n\pi}{3}} \]
Solve the quadratic trigonometric equation: \[ - 2 \sin^2 x - \cos x = - 1 \]
When an equation has both sine squared and cosine, use the Pythagorean identity \( \sin^2 x = 1 - \cos^2 x \) to convert everything to cosine.
\[ -2(1 - \cos^2 x) - \cos x = -1 \]
\[ -2 + 2\cos^2 x - \cos x = -1 \]
Move everything to one side to form a standard quadratic equation (\(ax^2+bx+c=0\)):
\[ 2 \cos^{2} x - \cos x - 1 = 0 \]
Factor the quadratic (treating \(\cos x\) like a variable \(u\)):
\[ (2 \cos x + 1)(\cos x - 1) = 0 \]
Solve each factor independently:
Case 1: \( 2 \cos x + 1 = 0 \Rightarrow \cos x = -\dfrac{1}{2} \)
\[ x = \mathbf{\dfrac{2\pi}{3} + 2n\pi} \quad \text{and} \quad x = \mathbf{\dfrac{4\pi}{3} + 2n\pi} \]
Case 2: \( \cos x - 1 = 0 \Rightarrow \cos x = 1 \)
Cosine is 1 at 0 radians.
\[ x = 0 + 2n\pi \Rightarrow \mathbf{x = 2n\pi} \]
Challenge 1 (Double Angle Identity): Solve for \(x\): \[ \sin(2x) + \cos(x) = 0 \]
Step 1: Apply the double angle identity for sine: \( \sin(2x) = 2\sin(x)\cos(x) \).
\[ 2\sin(x)\cos(x) + \cos(x) = 0 \]
Step 2: Factor out the common term, \( \cos(x) \).
\[ \cos(x)(2\sin(x) + 1) = 0 \]
Step 3: Set each factor to zero and solve.
Factor 1: \( \cos(x) = 0 \Rightarrow \mathbf{x = \dfrac{\pi}{2} + n\pi} \)
Factor 2: \( 2\sin(x) + 1 = 0 \Rightarrow \sin(x) = -\dfrac{1}{2} \)
\[ \Rightarrow \mathbf{x = \dfrac{7\pi}{6} + 2n\pi} \quad \text{and} \quad \mathbf{x = \dfrac{11\pi}{6} + 2n\pi} \]
Challenge 2 (Pythagorean Identity): Solve for \(x\): \[ \tan^2(x) - \sec(x) = 1 \]
Step 1: Use the identity \( \tan^2(x) = \sec^2(x) - 1 \) to get everything in terms of secant.
\[ (\sec^2(x) - 1) - \sec(x) = 1 \]
Step 2: Rearrange into a quadratic equation set to zero.
\[ \sec^2(x) - \sec(x) - 2 = 0 \]
Step 3: Factor the quadratic equation.
\[ (\sec(x) - 2)(\sec(x) + 1) = 0 \]
Step 4: Solve for cosine using reciprocal identities.
Factor 1: \( \sec(x) = 2 \Rightarrow \cos(x) = \dfrac{1}{2} \Rightarrow \mathbf{x = \dfrac{\pi}{3} + 2n\pi, \dfrac{5\pi}{3} + 2n\pi} \)
Factor 2: \( \sec(x) = -1 \Rightarrow \cos(x) = -1 \Rightarrow \mathbf{x = \pi + 2n\pi} \)
Challenge 3 (Extraneous Solutions): Solve for \(x\) on the interval \( [0, 2\pi) \): \[ \sin(x) - \cos(x) = 1 \]
Step 1: Square both sides of the equation. Warning: Squaring can introduce extraneous solutions, so you must check your final answers.
\[ (\sin(x) - \cos(x))^2 = 1^2 \]
\[ \sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = 1 \]
Step 2: Use the identities \( \sin^2(x) + \cos^2(x) = 1 \) and \( 2\sin(x)\cos(x) = \sin(2x) \).
\[ 1 - \sin(2x) = 1 \]
\[ \sin(2x) = 0 \]
Step 3: Solve for \( 2x \). Sine is zero at integer multiples of \( \pi \).
\[ 2x = n\pi \Rightarrow x = \dfrac{n\pi}{2} \]
On the interval \( [0, 2\pi) \), the possible angles are \( 0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} \).
Step 4: Check for extraneous solutions in the original equation \( \sin(x) - \cos(x) = 1 \).
Answer: \( \mathbf{x = \dfrac{\pi}{2}} \) and \( \mathbf{x = \pi} \)