# Solving trigonometric equations in degrees

## Example

Solve the equation $$\sin x^\circ = 0.5$$, where $$0 \le x \textless 360$$.

### Solution

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

Therefore, from the graph of the function, we can see that we should be expecting 2 solutions: 1 solution being between $$0^\circ$$ and $$90^\circ$$ and the other solution between $$90^\circ$$ and $$180^\circ$$.

$\sin x^\circ = 0.5$

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

$x^\circ = 30^\circ$

So we know that the first solution is $$30^\circ$$ as previously predicted from the graph.

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

Therefore since the trig equation we are solving is sin and it is positive (0.5), then we are in the 1st and 2nd quadrants.

We have already found the first solution which is the acute angle from the 1st quadrant, so to find the second solution, we need to use the rule in the 2nd quadrant.

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

$x^\circ = 150^\circ$

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

Question

Solve the equation $$\sin x^\circ = - 0.349$$, where $$0 \le x \textless 360$$.

From the graph of the function, we can see that we should be expecting 2 solutions: 1 solution between $$180^\circ$$ and $$270^\circ$$ and the other between $$270^\circ$$ and $$360^\circ$$.

$\sin x^\circ = - 0.349$

Since this is sin, but negative this means that we will be in the two quadrants where the sine function is negative - the third and fourth quadrants.

We need to firstly find the acute angle to use with the rules in these quadrants.

$\sin x^\circ = - 0.349$

When calculating the acute angle, we ignore the negative.

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

$x^\circ = 20.4261...$

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

$x^\circ = 180^\circ + 20.4^\circ$
$x^\circ = 200.4^\circ$
$x^\circ = 360^\circ - 20.4^\circ$
$x^\circ = 339.6^\circ$
Therefore $$x^\circ = 200.4^\circ ,\,339.6^\circ$$