Solving Trigonometric Equations

$2{\sin }^{2}a+3\sin a-1=0$

Step 1: Replace your trig function with a variable of your choice.

In our example, we will say let sin (a) = x

$2x^2+3x-1=0$

Step 2: Use the quadratic formula to solve for your variable.

$$\begin{gathered} a=2b=3c=-1 \\ x=\frac{-\left(3\right)\pm \sqrt{3^2-4\left(2\right)(-1)}}{2\left(2\right)} \\ =\frac{-3\pm \sqrt{17}}{4} \end{gathered}$$

Step 3: Replace your variable back as the function and take the inverse of the function to solve for the + equation. ( $\pm ,meansthereare2solutions$)

$$\begin{gathered} {\sin }^{-1}\left(\frac{-3+\sqrt{17}}{4}\right)=18.11° \\ x=\sin (18.11)=0.28 \end{gathered}$$

Step 4: Use the unit circle to determine the solution to the - equation as the domain of the inverse function is $\left[-1,1\right]$.

Due to sine being positive in the first and second quadrants, the second solution would be:

$$\begin{gathered} x=\mathrm{\pi }-0.28 \\ =2.86 \end{gathered}$$

$\cos x\cos \left(2x\right)+\sin x\sin \left(2x\right)=\frac{\sqrt{3}}{2}$

Step 1: Simplify your equation with a known identity.

In this example, it is the difference formula for cosine: $\cos a\cos b+\sin a\sin b=\cos (a-b)$

$$\begin{gathered} \cos x\cos \left(2x\right)+\sin x\sin \left(2x\right)=\frac{\sqrt{3}}{2} \\ c\mathrm{os}(x-2x)=\frac{\sqrt{3}}{2} \\ \cos (-x)=\frac{\sqrt{3}}{2} \\ \cos \left(x\right)=\frac{\sqrt{3}}{2},inthispartweusedthenegativeangletrigidentity:\cos (-x)=\cos \left(x\right) \end{gathered}$$

Step 2: Use the unit circle to determine the values of your angle (x).

In our example, we will focus on the 4th and 1st quadrants as cosine is positive in those quadrants.

Therefore, $x=\frac{\mathrm{\pi }}{6}andx=\frac{11\mathrm{\pi }}{6}$

$\cos \left(2x\right)=\frac{1}{2}on[0,2\mathrm{\pi })$

Step 1: Determine the quadrants of your initial solutions and the possible angles by using the unit circle.

$$\begin{gathered} {\cos }^{\_1}\left(\frac{1}{2}\right)=60° \\ \therefore possibleanglesare2x=\frac{\mathrm{\pi }}{3}and2x=\frac{5\mathrm{\pi }}{3} \end{gathered}$$

Step 2: Calculate the value of your initial solutions by dividing the possible angle by the number of angles.

$$\begin{gathered} 2x=\frac{\mathrm{\pi }}{3} \\ x=\frac{\mathrm{\pi }}{6} \\ 2x=\frac{5\mathrm{\pi }}{3} \\ x=\frac{5\mathrm{\pi }}{6} \end{gathered}$$

Step 3: Determine your other solutions by revolving around the circle by the number of angles and only selecting the answers within your range.

$$\begin{gathered} Firstquadrant2x=\frac{\mathrm{\pi }}{3} \\ \mathrm{First}\mathrm{rotation}:2x=\frac{\mathrm{\pi }}{3}+2\mathrm{\pi } \\ =\frac{7\mathrm{\pi }}{3} \\ \mathrm{x}=\frac{7\mathrm{\pi }}{6} \\ \mathrm{This}\mathrm{value}\mathrm{is}\mathrm{between}0\mathrm{and}2\mathrm{\pi }\mathrm{and}\mathrm{is}\mathrm{therefore}\mathrm{a}\mathrm{solution} \\ \mathrm{Second}\mathrm{Rotation}:2\mathrm{x}=\frac{\mathrm{\pi }}{3}+4\mathrm{\pi } \\ =\frac{13\mathrm{\pi }}{3} \\ \mathrm{x}=\frac{13\mathrm{\pi }}{6} \\ \mathrm{This}\mathrm{value}\mathrm{is}\mathrm{greater}\mathrm{than}2\mathrm{\pi }\mathrm{and}\mathrm{is}\mathrm{therefore}\mathrm{not}\mathrm{a}\mathrm{solution}. \\ \mathrm{Fourth}quadrant:2x=\frac{5\mathrm{\pi }}{3} \\ FirstRotation:2x=\frac{5\mathrm{\pi }}{3}+2\mathrm{\pi } \\ =\frac{11\mathrm{\pi }}{3} \\ \mathrm{x}=\frac{11\mathrm{\pi }}{6} \\ \mathrm{This}\mathrm{value}\mathrm{is}\mathrm{between}\mathrm{the}\mathrm{range}\mathrm{and}\mathrm{is}\mathrm{therefore}\mathrm{a}\mathrm{solution}. \\ \mathrm{Second}\mathrm{Rotation}:2\mathrm{x}=\frac{5\mathrm{\pi }}{3}+4\mathrm{\pi } \\ 2\mathrm{x}=\frac{17\mathrm{\pi }}{3} \\ \mathrm{x}=\frac{17\mathrm{\pi }}{6} \\ \mathrm{This}\mathrm{value}\mathrm{is}\mathrm{not}\mathrm{within}\mathrm{your}\mathrm{range}\mathrm{and}\mathrm{thefore}\mathrm{cant}\mathrm{be}\mathrm{a}\mathrm{solution}. \end{gathered}$$