By adjusting the given range, it is possible to solve equations of the form sin(nθ)=p, cos(nθ)=p and tan(nθ)=p. Similar adjustments can also be used to solve equations of the form sin(θ+α)=p, cos(θ+α)=p and tan(θ+α)=p.
Simple trigonometric equations with nθ
You have already learned that equations of the form sin(θ)=p and cos(θ)=p have solutions in a given range when −1≤p≤1. It follows that equations of the form sin(nθ)=p and cos(nθ)=p also have solutions when −1≤p≤1.
Now, consider finding the solutions to an equation of the form sin(nθ)=p, cos(nθ)=p or tan(nθ)=p over some interval D≤θ≤D′. Notice that if D≤θ≤D′ then it follows that nD≤nθ≤nD′. Therefore, equations with nθ can be solved by multipying up the range for θ given in the question. The final values for θ can then be found by dividing all the solutions in the range nD≤nθ≤nD′ by n to give D≤θ≤D′.
Example 1
Find all solutions to the equation sin(2θ)=21 in the interval 0°≤θ≤360°.
First multiply up the range. As the trig function has 2θ, multiply the range by 2 to give:
0°≤2θ≤720°
This means the solutions to sin(2θ)=21 for 2θ must lie in the range 0°≤2θ≤720°.
Find the principal value:
sin(2θ)2θ2θ=21=sin−1(21)=30°
Using a sin graph or otherwise, find that the full set of solutions in the interval 0°≤2θ≤720°:
2θ=30°,180−30150°,30+360390° and 150+360510°.
Dividing through by n=2, the full set of solutions to sin(2θ)=21 in the range 0°≤θ≤360° is given by:
Therefore, the full set of solutions to the equation sin(2θ)=21 in the interval 0°≤θ≤360°is given by θ=15°,75°,195°,255°.
Simple trigonometric equations with θ+α
Similar to the above, for D≤θ≤D′ then D+α≤θ+α≤D′+α. Therfore, when given equations of the form sin(θ+α)=p, cos(θ+α)=p or tan(θ+α)=p, find solutions in the range D+α≤θ+α≤D′+α before subtracting α. This will results in the range for θ being D≤θ≤D′.
Example 2
Find all solutions to the equation cos(θ+45°)=75 in the interval 0°≤θ≤360°. Give your answers to 1 decimal place.
First, adjust the range of θ by adding 45°to range 0°≤θ≤360° to give the interval 45°≤θ+45°≤405°.
Find the principal value:
cos(θ+45°)θ+45°θ+45°=75=cos−1(75)=44.415…°
This lies outside the desired range! Using a cos graph or otherwise, deduce that the full set of solutions lying in the range 45°≤θ+45°≤405° is given by:
Therefore, the full set of solutions to the equation cos(θ+45°)=75, to 1 decimal place, in the interval 0°≤θ≤360°, is given by θ=270.6° and θ=359.4°.
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FAQs - Frequently Asked Questions
How do I solve equations of the form sin(x + a) = p?
Solve sin(x + a) = p in the range D+a< x+a < D'+a, then subtract a from all the solutions to give all the values in the range D < x < D'.
How do I solve equations of the form sin(nx) = p?
Find the solutions for sin(nx) = p in the range nD < nx < nD'. Then divide all the solutions by n to give all the solutions in the range D < x < D'.
