Trigonometry: Trigonometric functions
Trigonometric equations 1
We have seen that the trigonometric functions #\sin(x)#, #\cos(x)# and #\tan(x)# on a limited domain have an inverse function. With this inverse function we can solve equations of the form #\sin(x)=c#, #\cos(x)=c# and #\tan(x)=c#. For limited values of #c# we can give exact solutions, for other values we do this with the aid of the calculator.
Solving trigonometric equations
Stepbystep 
Example 

We solve a trigonometric equation of the form #\sin(ax+b)=c#, #\cos(ax+b)=c# or #\tan(ax+b)=c# with numbers #a#, #b# and #c#. 
#\sin(2x+\pi)=\frac{1}{\sqrt{2}}# 

Step 1 
Use the inverse function to determine one solution of #ax+b#. This solution is repeated every period; we indicate this by adding #k \cdot 2 \pi#, in which #k# is an integer. Note: the calculator does not give exact solutions. If an exact solution is required, we must use the table with special values. 
#2x+\pi=\frac{\pi}{4}+k \cdot 2\pi# 
Step 2 
Use the graphs of the trigonometric functions or the symmetry of the unit circle to determine if there is a second solution within one period. This solution is also repeated every period. 
#\begin{array}{c}2x+\pi=\frac{\pi}{4}+k \cdot 2\pi \\ \lor \\ 2x+\pi=\frac{3\pi}{4}+k\cdot 2\pi\end{array}# 
Step 3 
Reduce the left side of the equation to #x#. 
#\begin{array}{c}x=\frac{3\pi}{8}+k \cdot \pi \\ \lor \\x=\frac{\pi}{8}+k\cdot \pi\end{array}# 
#\begin{array}{rcl}
\cos({{\pi}\over{2}}+{{x}\over{2}})&=&{{\sqrt{3}}\over{2}} \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}}\\
{{\pi}\over{2}}+{{x}\over{2}}={{\pi}\over{6}}+2\cdot \pi\cdot k &\lor& {{\pi}\over{2}}+{{x}\over{2}}={{11\cdot \pi}\over{6}}+2\cdot \pi\cdot k \\ &&\phantom{xxx}\blue{\text{taken the inverse }\cos \text{ on both sides}}\\
x={{4\cdot \pi}\over{3}}+4\cdot \pi\cdot k &\lor& x={{14\cdot \pi}\over{3}}+4\cdot \pi\cdot k \\ &&\phantom{xxx}\blue{\text{reduced}}\\
\end{array}#
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