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Question: How do you solve \({{\cos }^{3}}\left( x \right)=\cos \left( x \right)\) on the interval of \([0,2\p...

How do you solve cos⁡3(x)=cos⁡(x){{\cos }^{3}}\left( x \right)=\cos \left( x \right) on the interval of [0,2π)?[0,2\pi )?

Explanation

Solution

Here a trigonometric equation is given with its interval which we have to simplify it.
Here, We are simply substituting the other variable ′u′'u' at the place of cos⁡x\cos x
We did it to make the equation more simple.
After simplifying it we will get the value for the variable ′u′'u' which is nothing but the value of cos⁡x\cos x put that all, values in cos⁡x\cos x and we will get the final answer in terms of angles.

Complete step by step solution:
Given that, there is an equation,
cos⁡3(x)=cos⁡(x)...(i){{\cos }^{3}}\left( x \right)=\cos \left( x \right)...(i)
Whose interval is [0,2π)[0,2\pi ) and we have to solve it. Here, we have to use a substitution method that is putting a variable at the place of cos⁡(x)\cos \left( x \right) and then we have to solve it in the way of a normal regular polynomial.
cos⁡3(x)=cos⁡(x){{\cos }^{3}}\left( x \right)=\cos \left( x \right)
(cos⁡x)3=cos⁡(x){{\left( \cos x \right)}^{3}}=\cos \left( x \right)
Putting cos⁡x=u\cos x=u in above equation, we get
u3=u{{u}^{3}}=u
Transpose uu at the left side to form an equation.
u3−u=0{{u}^{3}}-u=0
u(u2−1)=0u\left( {{u}^{2}}-1 \right)=0
u(u−1)(u+1)=0u\left( u-1 \right)\left( u+1 \right)=0
Separate the above terms, such that,
u=0u=0
u−1=0u-1=0, u+1=0u+1=0
u=1u=1, u=−1u=-1
The values of u=0,1,−1u=0,1,-1
cos⁡x=0,1,−1\cos x=0,1,-1
Now, put this value in cos⁡x\cos x i.e. in cosine equation i.e. put cos⁡x=0\cos x=0
x=cos⁡−1(0)x={{\cos }^{-1}}\left( 0 \right)
x=π2,3π2x=\dfrac{\pi }{2},\dfrac{3\pi }{2}
Now, put cos⁡x=1\cos x=1
x=cos⁡−1(1)x={{\cos }^{-1}}\left( 1 \right)
x=0x=0
Now, put cos⁡x=1\cos x=1
x=cos⁡−1(1)x={{\cos }^{-1}}\left( 1 \right)
x=0x=0
Now, put cos⁡x=−1\cos x=-1
x=cos⁡−1(−1)x={{\cos }^{-1}}\left( -1 \right)
x=πx=\pi

Therefore, the final answer for the given trigonometric equation is x=0,π,π2,3π2x=0,\pi ,\dfrac{\pi }{2},\dfrac{3\pi }{2}.

Note: In this question, the trigonometric equation is given which we have to solve.
Here we put cos⁡x=u\cos x=u just for simplifying the equation. Otherwise using trigonometric functions the equation may become more complicated.
The final answer we got is nothing but the trigonometric cosine values in degree.
i.e. cos⁡−1(0)=π2,3π2{{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{2},\dfrac{3\pi }{2} i.e. 90∘,270∘90{}^\circ ,270{}^\circ
cos⁡−1(1)=0{{\cos }^{-1}}\left( 1 \right)=0 i.e. 0∘0{}^\circ
cos⁡−1(−1)π{{\cos }^{-1}}\left( -1 \right)\pi i.e. 180∘180{}^\circ