Trigonometric relationships

Solving trigonometric equations

Many students think they've solved a trig equation when they get one answer (one size of angle x^\circ). However there's often more than one answer expected so be aware of this.

Example

Solve the equation \sin x^\circ  = 0.5, where 0 ≤ x < 360.

Answer

Let's remind ourselves of what the sine graph looks like so that we can see how many solutions we should be expecting:

Diagram of a sin equation graph

Therefore, from the graph of the function, we can see that we should be expecting two solutions: one solution being between 0˚ and 90˚ and the other solution between 90˚ and 180˚.

\sin x^\circ  = 0.5

x^\circ  = {\sin ^{ - 1}}(0.5)

x^\circ  = 30^\circ

So we know that the first solution is 30˚ as previously predicted from the graph.

To get the other solution, we need to go back to our quadrants and use the appropriate rule:

Diagram of quadrant rules

Therefore since the trig equation we are solving is sin and it is positive (0.5), then we are in the first and second quadrants. We have already found the first solution which is the acute angle from the first quadrant, so to find the second solution, we need to use the rule in the second quadrant.

x^\circ  = 180^\circ  - 30^\circ

x^\circ  = 150^\circ

Therefore x^\circ  = 30^\circ ,150^\circ

Now try the example questions below.

Question

Solve the equation \sin x^\circ  = -0.349, where 0 ≤ x < 360.

Diagram of a sin graph with equation y = sin x° (y= -0.349)

From the graph of the function, we can see that we should be expecting two solutions: one solution between 180° and 270° and the other between 270° and 360°.

\sin x^\circ  =  - 0.349

Since this is sin, but negative, this means that we will be in the two quadrants where the sine function is negative, ie the third and fourth quadrants. We need to firstly find the acute angle to use with the rules in these quadrants.

When calculating the acute angle, we ignore the negative.

x^\circ  = {\sin ^{ - 1}}(0.349)

x^\circ  = 20.4^\circ \,(to\,1d.p.)

Third quadrant

x^\circ  = 180^\circ  + 20.4^\circ

x^\circ  = 200.4^\circ

Fourth quadrant

x^\circ  = 360^\circ  - 20.4^\circ

x^\circ  = 339.6^\circ

Therefore: x^{\circ}= 200.4^{\circ}, 339.6^{\circ}

Question

Solve the equation 4\sin x^\circ  - 3 = 0, where 0 \le x \textless 360.

When solving a question like this one, we need to rearrange it first.

4\sin x^\circ  - 3 = 0

4\sin x^\circ  = 0 + 3

4\sin x^\circ  = 3

\sin x^\circ  = \frac{3}{4}

The graph of this function looks like this:

Diagram of a sin graph with equation y = sin x° (y= 0.75)

From the graph of the function, we can see that we should be expecting two solutions: one solution between 0° and 90° and the other between 90° and 180°.

\sin x^\circ  = \frac{3}{4}

Since this is sin and is positive this means that we will be in the two quadrants in which the sine function is positive, ie the first and second quadrants.

First quadrant

\sin x^\circ  = \frac{3}{4}

x^\circ  = {\sin ^{ - 1}}(\frac{3}{4})

x=sin^{-1}0.75

x^\circ  = 48.6^\circ (to\,1\,d.p.)

Second quadrant

x^\circ  = 180^\circ  - 48.6^\circ

x^\circ  = 131.4^\circ

Therefore x^\circ  = 48.6^\circ ,131.4^\circ