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Solving trigonometric equations

Solving trigonometric equations

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Tutor: Meera

Summary

Solving trigonometric equations

​​In a nutshell

You have already learnt how to solve trigonometric equations in degrees. Trigonometric equations can be solved in radians using a similar method.



Change angle units on a calculator

You can solve trigonometric equations as you have already done, however make sure to set a calculator in 'radians' mode as follows:

Maths; Radians; KS5 Year 13; Solving trigonometric equations

CALCULATOR TIP


Select 'Shift' and 'SETUP'. The display should show:

1:Input/Output2:Angle Unit3:Number format4:Engineer Symbol\boxed{\begin{aligned}1&: Input/Output \\2&: Angle \ Unit \\3&: Number \ format \\4&: Engineer \ Symbol\end{aligned}}​​


Select option 22 to change the angle units. The display will show:

1:Degree2:Radian3:Gradian\boxed{\begin{aligned}1&: Degree \\2&: Radian \\3&: Gradian\end{aligned}}​​


Select either 11 for degrees or 22 for radians.​



Solve trigonometric equations

Trigonometric equations in radians can now be solved in the same way as solving equations in degrees. Once you have calculated the 'principle value' after using the inverse trig function on a calculator, find the other values in the range either by using the appropriate trigonometric graph with the angles on the xx-axis given in radians, or by using the following to work out the other values:


trig equation

principal value

second value

other values

sin(θ)=k\sin(\theta) = k​​
θ\theta​​
πθ\pi - \theta​​
Add or subtract 2π2 \pi to the principal and second values according to the given range.​
cos(θ)=k\cos(\theta) = k​​
θ\theta​​
2πθ2 \pi - \theta​​
tan(θ)=k\tan(\theta) = k​​
θ\theta​​
θ+π\theta + \pi​​


Note: There are usually two answers between 00 and 2π2 \pi radians. The first answer is the principal value from a calculator, the second value can be found by using the symmetry of a trigonometric graph (used in the table above). Any other values can be found by taking the two answers and either adding or subtracting 2π2 \pi from the first two answers.


Example 1

Solve 2sin3θ=12 \sin 3\theta = 1. Give all answers in the range 0θ2π0 \le \theta \le 2 \pi.


Rearrange for sin3θ\sin 3 \theta and solve:

2sin(3θ)=1sin(3θ)=123θ=sin1(12)3θ=π6\begin{aligned}2 \sin (3 \theta) &= 1 \\\\\sin (3 \theta) &= \dfrac 1 2 \\\\ 3 \theta &= \sin^{-1} \Big( \dfrac 1 2 \Big ) \\\\3 \theta &= \dfrac {\pi}{6}\end {aligned}​​


This gives the principal value of π6\dfrac {\pi}{6}. Adjust the range given and work out the other values. 


The range for θ\theta  is given as:

0θ2π0 \le \theta \le 2 \pi​​


Adjust the range for 3θ3 \theta:

03θ6π0 \le 3\theta \le 6 \pi​​


Working out all values of 3θ3 \theta up to 6π6 \pi using a graph or alternative method:

Maths; Radians; KS5 Year 13; Solving trigonometric equations

3θ=π6,5π6ππ6,13π6π6+2π,17π65π6+2π,25π613π6+2π,29π617π6+2π3 \theta = \dfrac {\pi}{6 }, \underbrace{\dfrac {5 \pi}{6 }}_{\pi - \frac {\pi}{6}}, \underbrace{\dfrac {13 \pi}{6 }}_{\frac {\pi}{6 }+2 \pi}, \underbrace{\dfrac {17 \pi}{6 }}_{\frac {5 \pi}{6 }+2 \pi}, \underbrace{\dfrac {25 \pi}{6 }}_{\frac {13 \pi}{6 }+2 \pi}, \underbrace {\dfrac {29 \pi}{6 }}_{\frac {17 \pi}{6 }+2 \pi}​​


Divide by 33 to find all values of θ\theta

θ=π18,5π18,13π18,17π18,25π18,29π18 \underline{\theta = \dfrac {\pi}{18 }, \dfrac {5 \pi}{18 }, \dfrac {13 \pi}{18 }, \dfrac {17 \pi}{18 }, \dfrac {25 \pi}{18 }, \dfrac {29 \pi}{18 }}​​



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