Trigonometric equations 2

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#.

Solving trigonometric equations of the form sin(A)=sin(B)

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

\[\blue{x+\pi}=\green{2x}+k\cdot2 \pi \lor \blue{x+\pi}=\pi-\green{2x}+k \cdot 2 \pi\]

Solving further

The obtained solutions still need to be reduced further before on the left side of the equations only #x# remains. We do this using the normal rules for reduction.

Domain

When asked about solutions within a certain domain, it must be determined at the end for which value of #k# the solutions are within the domain.

Solving trigonometric solutions of the form cos(A)=cos(B)

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

\[\blue{x+\pi}=\green{2x}+k\cdot2 \pi \lor \blue{x+\pi}=-\green{2x}+k \cdot 2 \pi\]

Solving further

The obtained solutions still need to be reduced further before on the left side of the equations only #x# remains. We do this using the normal rules for reduction.

Domain

When asked about solutions within a certain domain, it must be determined at the end for which value of #k# the solutions are within the domain.

Solving trigonometric equations of the form tan(A)=tan(B)

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

\[\blue{x+\pi}=\green{2x}+k\cdot \pi\]

Solving further

The obtained solutions still need to be reduced further before on the left side of the equations only #x# remains. We do this using the normal rules for reduction.

Domain

When asked about solutions within a certain domain, it must be determined at the end for which value of #k# the solutions are within the domain.

Solve the equation \[\sin\left({{x}\over{4}}\right)=\sin\left({{\pi}\over{4}}+{{x}\over{2}}\right)\].
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.

#x=-\pi-8\cdot \pi\cdot k \lor x=\pi+{{8\cdot \pi\cdot k}\over{3}}#

#\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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