Trigonometric Equations (Grade 11 NSC Matric Mathematics): Revision Notes

Worked Example: Solving tan θ = 5

Question: Solve for $\theta$ (correct to one decimal place), given $\tan \theta = 5$ and $\theta \in [0°; 360°]$.

Solution:

Step 1: Find the reference angle $\tan \theta = 5$ $\therefore \theta = \tan^{-1} 5 = 78.7°$

This is an acute angle in the first quadrant (the reference angle).

Step 2: Use CAST diagram for tangent The CAST diagram shows that $\tan \theta$ is positive in the first and third quadrants.

Step 3: Find angles in correct quadrants

• First quadrant: $\theta = 78.7°$
• Third quadrant: $\theta = 180° + 78.7° = 258.7°$

Step 4: Check solutions using calculator

Final answer: $\theta = 78.7°$ or $\theta = 258.7°$

Worked Example: Solving cos α = -0.7

Question: Solve for $\alpha$ (correct to one decimal place), given $\cos \alpha = -0.7$ and $\theta \in [0°; 360°]$.

Solution:

Step 1: Find the reference angle We use the positive value: $\text{ref } \angle = \cos^{-1} 0.7 = 45.6°$

Step 2: Use CAST diagram for cosine Cosine is negative in the second and third quadrants.

Step 3: Find angles in correct quadrants

• Second quadrant: $\alpha = 180° - 45.6° = 134.4°$
• Third quadrant: $\alpha = 180° + 45.6° = 225.6°$

Final answer: $\alpha = 134.4°$ or $\alpha = 225.6°$

Worked Example: General solution for sin θ = 0.3

Question: Determine the general solution for $\sin \theta = 0.3$.

Solution:

Step 1: Find the reference angle $\text{ref } \angle = \sin^{-1} 0.3 = 17.5°$

Step 2: Use CAST diagram Sine is positive in the first and second quadrants.

Step 3: Find basic solutions

• First quadrant: $\theta = 17.5°$
• Second quadrant: $\theta = 180° - 17.5° = 162.5°$

Step 4: Add multiples of the period ($360°$)

General solution:

$\theta = 17.5° + k \times 360°$ or $\theta = 162.5° + k \times 360°$, where $k \in \mathbb{Z}$

Worked Example: General solution for cos 2θ = -0.6427

Question: Determine the general solution for $\cos 2\theta = -0.6427$.

Solution:

Step 1: Find the reference angle $\text{ref } \angle = \cos^{-1} 0.6427 = 50.0°$

Step 2: Use CAST diagram Cosine is negative in the second and third quadrants.

Step 3: Find solutions for the angle $2\theta$

• Second quadrant: $2\theta = 180° - 50° = 130°$
• Third quadrant: $2\theta = 180° + 50° = 230°$

Step 4: Solve for $\theta$

• $\theta = 65° + k \times 180°$
• $\theta = 115° + k \times 180°$

Important: Remember to divide the period ($360°$) by the coefficient of $\theta$.

General solution: $\theta = 65° + k \times 180°$ or $\theta = 115° + k \times 180°$, where $k \in \mathbb{Z}$

Worked Example: Solving 4sin²θ = 3

Question: Find the general solution of $4\sin^2\theta = 3$.

Solution:

Step 1: Simplify the equation $4\sin^2\theta = 3$ $\sin^2\theta = \frac{3}{4}$ $\sin \theta = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}$

Step 2: Find reference angle $\text{ref } \angle = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60°$

Step 3: Determine quadrants

• For $\sin \theta = +\frac{\sqrt{3}}{2}$: first and second quadrants
• For $\sin \theta = -\frac{\sqrt{3}}{2}$: third and fourth quadrants

Step 4: Find all solutions

• $\theta = 60° + k \times 360°$ (first quadrant)
• $\theta = 120° + k \times 360°$ (second quadrant)
• $\theta = 240° + k \times 360°$ (third quadrant)
• $\theta = 300° + k \times 360°$ (fourth quadrant)

General solution: $\theta = 60° + k \times 360°$, $\theta = 120° + k \times 360°$, $\theta = 240° + k \times 360°$, or $\theta = 300° + k \times 360°$, where $k \in \mathbb{Z}$

Worked Example: Solving 2cos²θ - cos θ - 1 = 0

Question: Find $\theta$ if $2\cos^2\theta - \cos \theta - 1 = 0$ for $\theta \in [-180°; 180°]$.

Solution:

Step 1: Factor the equation

• $2\cos^2\theta - \cos \theta - 1 = 0$
• $(2\cos \theta + 1)(\cos \theta - 1) = 0$

Step 2: Solve each factor

Either $2\cos \theta + 1 = 0$ or $\cos \theta - 1 = 0$

From $2\cos \theta + 1 = 0$:

• $\cos \theta = -\frac{1}{2}$
• $\text{ref } \angle = 60°$

Solutions: $\theta = 120°$ and $\theta = 240°$ (or $\theta = -120°$ within the given interval)

From $\cos \theta - 1 = 0$:

• $\cos \theta = 1$
• $\theta = 0°$

Step 3: Check which solutions lie in $[-180°; 180°]$

• $\theta = -120°, 0°, 120°$

Final answer: $\theta = -120°, 0°, 120°$