Linear Trigonometric Equations (AQA A-Level Mathematics): Revision Notes

Example 1: Solving $\sin \theta$ = 0.5

• Step 1: Isolate the function: $\sin \theta = 0.5$

• Step 2: Find the principal solution: $\theta = \sin^{-1}(0.5)$ The principal solution is $\theta = 30^\circ (since \sin 30^\circ = 0.5).$

• Step 3: Consider all solutions: $\theta = 30^\circ \quad \text{or} \quad \theta = 180^\circ - 30^\circ = 150^\circ$ (Because $\sin$ is positive in the first and second quadrants.)

• Step 4: List all solutions within the interval $\ 0^\circ \ to \ 360^\circ:$ $\theta = 30^\circ, 150^\circ$

Example 2: Solving $\cos \theta = -\frac{1}{2}$

• Step 1: Isolate the function: $\cos \theta = -\frac{1}{2}$

• Step 2: Find the principal solution: $\theta = \cos^{-1}\left(-\frac{1}{2}\right)$ The principal solution is $\theta = 120^\circ (since \cos 120^\circ = -\frac{1}{2}).$

• Step 3: Consider all solutions: $\theta = 120^\circ \quad \text{or} \quad \theta = 360^\circ - 120^\circ = 240^\circ$ (Because $\ \cos$ is negative in the second and third quadrants.)

• Step 4: List all solutions within the interval$\ 0^\circ\ to \ 360^\circ:$ $\theta = 120^\circ, 240^\circ$

Example 3: Solving $2\sin \theta + 1 = 0$

• Step 1: Isolate the function: $2\sin \theta = -1 \quad \Rightarrow \quad \sin \theta = -\frac{1}{2}$

• Step 2: Find the principal solution: $\theta = \sin^{-1}\left(-\frac{1}{2}\right)$ The principal solution is $\theta = -30^\circ, but \ to \ stay\ within\ the \ 0^\circ\ to \ 360^\circ$range, we convert this to$\ 330^\circ (since \ \sin(360^\circ - 30^\circ = -\frac{1}{2}.)$

• Step 3: Consider all solutions: $\theta = 180^\circ + 30^\circ = 210^\circ \quad \text{and} \quad \theta = 330^\circ$ (Because $\sin$ is negative in the third and fourth quadrants.)

• Step 4: List all solutions within the interval $\ 0^\circ\ to \ 360^\circ:$ $\theta = 210^\circ, 330^\circ$

Example 4: Solving $\tan \theta = 1$

• Step 1: Isolate the function: $\tan \theta = 1$

• Step 2: Find the principal solution: $\theta = \tan^{-1}(1)$ The principal solution is $\theta = 45^\circ (since \tan 45^\circ = 1).$

• Step 3: Consider all solutions: $\theta = 45^\circ + 180^\circ n$ (Because the period of $\ \tan\ is \ 180^\circ.)$

• Step 4: List all solutions within the interval $\ 0^\circ\ to \ 360^\circ:$ $\theta = 45^\circ, 225^\circ$

Example: Solve $\sin x = -0.15$ for $0 \leq x \leq 450$

Visual representation of $\sin x = -0.15$ on the graph with solutions indicated.

Solving $\cos(2x) = 0.2$ for $0 \leq x \leq 360$

$\xcancel{\cos x=0.1 (\div 2}$ The only think which can penetrate is cos bracket is $cos^{-1}$

Example

Solve $\tan(3x) = 0.27$ for $-180 \leq x \leq 180$

1. Rearrange the equation to solve for $3x$:

Extend the interval to accommodate the factor of $3$:

2. List out the angles:

3. Divide by $3$ to find $x$:

Solve $\sin(2x + 15) = \frac{\sqrt{2}}{2}$ for $0 \leq x \leq 180$

Example:

4. Find the first solution on your calculator:

Note: This is not required as a line of working, but it's part of the thought process.

Implied truncation rather than rounding.

• The calculation is done using a calculator, displaying $\cos^{-1}(0.2) = 78.46304097$.

Steps to Solve Trigonometric Equations:

1. Sketch the relevant graph within the domain given to identify other solutions.

   

2. Conclude with all valid solutions:

• $\boxed {x = 78.46^\circ, 281.5^\circ \quad \text{(4sf)}}$

Example:

Solve $\tan(x) = -0.3$, for $0 \leq x \leq 540$:

• Initial solution: $x = \arctan(-0.3) = -16.699^\circ$

• Note: The initial solution is invalid, but if the graph is extended, it can be used to find valid solutions.

  

• Mark the graph and use extensions to find solutions:

• $x = 163.3^\circ, 343.3^\circ, 523.3^\circ$