Trigonometry: Trigonometric functions
Trigonometric equations 2
We have seen how we can solve trigonometric equations of the form #\sin(ax+b)=c#, #\cos(ax+b)=c# and #\tan(ax+b)=c#. There is still another form of trigonometric equations we can easily solve, namely #\sin(A)=\sin(B)#, #\cos(A)=\cos(B)# and #\tan(A)=\tan(B)#, in which #A# and #B# are expressions in #x#.
The solution of the equation #\sin(\blue A)=\sin(\green B)#, in which #\blue A# and #\green B# are expressions in #x# has the following solutions:
\[\blue A=\green B+k \cdot 2 \pi \lor \blue A=\pi-\green B+k\cdot 2\pi\]
Here, #k# is an integer.
Example
\[\sin(\blue{x+\pi})=\sin(\green{2x})\]
has solutions
\[\blue{x+\pi}=\green{2x}+k\cdot2 \pi \lor \blue{x+\pi}=\pi-\green{2x}+k \cdot 2 \pi\]
The solution of the equation #\cos(\blue A)=\cos(\green B)#, in which #\blue A# and #\green B# are expressions in #x# has the following solutions:
\[\blue A=\green B+k \cdot 2 \pi \lor \blue A=-\green B+k\cdot 2\pi\]
Here, #k# is an integer.
Example
\[\cos(\blue{x+\pi})=\cos(\green{2x})\]
has solutions
\[\blue{x+\pi}=\green{2x}+k\cdot2 \pi \lor \blue{x+\pi}=-\green{2x}+k \cdot 2 \pi\]
The solution of the equation #\tan(\blue A)=\tan(\green B)#, in which #\blue A# and #\green B# are expressions in #x#, has the following solutions:
\[\blue A=\green B+k \cdot \pi \]
Here, #k# is an integer.
Example
\[\tan(\blue{x+\pi})=\tan(\green{2x})\]
has solutions
\[\blue{x+\pi}=\green{2x}+k\cdot \pi\]
Give your answer in the form: #x=x_1 \lor x=x_2 \lor \ldots x=x_n#, in which #x_1#, #x_2#, #\ldots#, #x_n# are the correct solutions and #n# is the number of solutions. Use #k# as a random integer.
#\begin{array}{rcl}
\sin\left({{x}\over{4}}\right)&=&\sin\left({{\pi}\over{4}}+{{x}\over{2}}\right) \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}}\\
{{x}\over{4}}={{\pi}\over{4}}+{{x}\over{2}}+2 \pi \cdot k &\lor& {{x}\over{4}}=\pi-({{\pi}\over{4}}+{{x}\over{2}})+2 \pi \cdot k \\ &&\phantom{xxx}\blue{\text{general rule for solution }\sin(A)=\sin(B)}\\
x=-\pi-8\cdot \pi\cdot k &\lor& x=\pi+{{8\cdot \pi\cdot k}\over{3}}\\ &&\phantom{xxx}\blue{\text{reduced}}\\
\end{array}#
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