Solving trigonometric equations in radians (extension)
To solve trigonometric equations involving radian measure, you should proceed in the same way as for degrees. Use the diagram below to help you apply the necessary conditions.
You should also know your exact values from the exact value triangles.
Example 1
Solve \(2\cos (x) = 1\), for \(0 \le x \le 2\pi\).
Solution
\(2\cos (x) = 1\)
\(\cos (x) = \frac{1}{2}\)
\(x = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)\)
Solve in the same way as you would have had it been in degrees.
As with equations in degrees, decide on the appropriate quadrants and then apply the necessary conditions.
\(x = \frac{\pi }{3}\)
Second quadrant
\(x = 2\pi - \frac{\pi }{3}\)
\(x = \frac{{5\pi }}{3}\)
Therefore \(x = \frac{\pi }{3},\,\frac{{5\pi }}{3}\)
Example 2
Solve \(\cos 2x + 3\sin x + 1 = 0\), for \(0 \textless x \textless 2\pi\).
Solution
\(\cos 2x + 3\sin x + 1 = 0\)
\(1 - 2{\sin ^2}x + 3\sin x + 1 = 0\)
\(- 2{\sin ^2}x + 3\sin x + 2 = 0\)
\(2{\sin ^2}x - 3\sin x - 2 = 0\)
\((2\sin x + 1)(\sin x - 2) = 0\)
\(2\sin x + 1 = 0\)
\(\sin x = - \frac{1}{2}\)
Since sin is negative, then we are in the third and fourth quadrants.
\(x = {\sin ^{ - 1}}\left( {\frac{1}{2}} \right)\)
\(x = \frac{\pi }{6}\)
Third quadrant
\(x = \pi + \frac{\pi }{6}\)
\(x = \frac{{7\pi }}{6}\)
Fourth quadrant
\(x = 2\pi - \frac{\pi }{6}\)
\(x = \frac{{11\pi }}{6}\)
\(\sin x - 2 = 0\)
\(\sin x = 2\)
Since \(0 \textless x \textless 2\pi\) then there are no solutions.
Therefore \(x = \frac{{7\pi }}{6},\,\frac{{11\pi }}{6}\)