7.07 Solving trigonometric equations 2
Trigonometric equations become more difficult to solve when compound angles are introduced or we are required to use our knowledge of trigonometric identities to manipulate them into a form that we can solve. Factorising becomes a key skill here, allowing us to solve trigonometric equations that involve more than one trigonometric ratio.
Remember: after any algebraic manipulation, we only ever use a positive value in our calculator to find the relative acute angle that solves the equation. We then draw a quadrant (ASTC) diagram to locate the values we need, and finally, check that our answers fit in the domain specified in the question.
In this set, we will consider another five types of trigonometric equations you might need to solve.
Harder types of trigonometric equations
Type 1: simple compound angle equations
When we need to find the size of an angle that has another value added or subtracted to the variable, the simplest way to proceed is to make a substitution for the whole compound angle and solve this simpler equation. (You can use any letter you wish, many people choose $u$u.) At the end of the problem, we can substitute back to the original form and solve this for the variable required. Make sure that your solutions fit in the original domain. If they don't, add or subtract $360^\circ$360° until they do.
Practice question
Question 1
We want to solve $\cos\left(\theta-100^\circ\right)=\frac{1}{2}$cos(θ−100°)=12 for $0^\circ\le\theta\le360^\circ$0°≤θ≤360°.
Let $\alpha=\theta-100^\circ$α=θ−100°. Hence, solve $\cos\alpha=\frac{1}{2}$cosα=12 for $-100^\circ\le\alpha\le260^\circ$−100°≤α≤260°.
Write your solutions on the same line, separated by a comma.
Hence, solve $\cos\left(\theta-100^\circ\right)=\frac{1}{2}$cos(θ−100°)=12 for $0^\circ\le\theta\le360^\circ$0°≤θ≤360°.
Write your solutions on the same line, separated by a comma.
Type 2: basic squared trigonometric functions
Taking square roots of both sides will turn this equation into a simple one with only one trigonometric function. Don't forget that both the positive and negative square roots are possible, so this type of question will result in answers in every quadrant within the specified domain.
Practice question
Question 2
Find the measure in degrees of the angles satisfying $2\sin^2\left(\theta\right)=1$2sin2(θ)=1 for $0^\circ\le\theta\le360^\circ$0°≤θ≤360°.
Type 3: equations that can be factorised
We can never cancel trigonometric functions that appear on both sides of an equation, but we can factorise! Trigonometric equations might have common factors. They can also be factorised as quadratics and the null factor law can be applied to find various solutions. Sometimes, these component factors might not be able to be solved as they involve a value that lies outside the range of sine or cosine, which are the values between $-1$−1 and $1$1 inclusive.
Practice question
Question 3
Find the measure in degrees of the angles satisfying $7\cos^2\left(\theta\right)+3\cos\theta=3$7cos2(θ)+3cosθ=3 for $0^\circ\le\theta\le360^\circ$0°≤θ≤360°. Give your answers correct to one decimal place.
Type 4: equations that can be simplified using trigonometric identities
If an equation has more than one trigonometric function in it and you can't factorise it into a form that you can solve, you will need to manipulate it into a different form using trigonometric identities. The Pythagorean identities and reciprocal ratios are commonly used in these difficult scenarios.
Practice question
Question 4
Solve the equation $\tan^2\left(\theta\right)+5=3\sec^2\left(\theta\right)$tan2(θ)+5=3sec2(θ) for $0^\circ\le\theta\le360^\circ$0°≤θ≤360°.
Write your solutions on the same line, separated by commas.
Type 5: harder compound angle equations involving multiples of $x$x or $\theta$θ
Like the simple compound angles discussed above, we can make a substitution to simplify the equation. These questions involve an additional step however as the domain given is generally written for a single variable and we need to change it to match the question.
For example, the equation $\tan3x=1,0^\circ\le x\le360^\circ$tan3x=1,0°≤x≤360° actually needs to be solved over the domain $0^\circ\le3x\le1080^\circ$0°≤3x≤1080° . This means we will have more than the usual two solutions! You might think that we will end up with solutions far too big for the domain. But the last step for these problems is to solve for $x$x, and so, when we divide the large solutions we have, in this case by $3$3, they will fall back into the original domain of $0^\circ\le x\le360^\circ$0°≤x≤360°.
Practice question
Question 5
Find the measure in degrees of the angles satisfying $2\sin3\theta-\sqrt{2}=0$2sin3θ−√2=0 for $0^\circ\le\theta\le360^\circ$0°≤θ≤360°.