How do you solve cos ⁡ ( 3 x ) = cos ⁡ ( x )?I have tried to rewrite it to 4cos3(x)−3cos(x)=cos(x) and then solve it. Add 3cos(x) to both sides and then divide by 4. And then I have this: cos3(x)=cos(x) . I don’t know what so do now or even if on the right track.

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How do you solve   cos  ⁡  (  3  x  )  =  cos  ⁡  (  x  )?I have tried to rewrite it to 4cos3(x)−3cos(x)=cos(x) and then solve it. Add 3cos(x) to both sides and then divide by 4. And then I have this: cos3(x)=cos(x) . I don’t know what so do now or even if on the right track.

merlatas497 · Accepted Answer

As an alternative (and cleaner) solution, you can note that two angles   θ  ,      θ    ′   have the same cosine if and only if   θ  ≡      θ    ′      (  mod    2  π  ) or   θ  ≡  −      θ    ′      (  mod    2  π  ). In your case, you thus have:                    cos        ⁡        (        3        x        )        =        cos        ⁡        (        x        )                                    ⟺                3        x        ≡        x                (        mod                2        π        )        ,         or         3        x        ≡        −        x                (        mod                2        π        )                                                  ⟺                2        x        ≡        0                (        mod                2        π        )        ,         or         4        x        ≡        0                (        mod                2        π        )                                                  ⟺                x        ≡        0                (        mod                π        )        ,         or         x        ≡        0                (        mod                π                  /                2        )                                                  ⟺                x        ≡        0                (        mod                π                  /                2        )            where, for the last equivalence, we use that if   x  =  k  π for some integer k, then of course   x  =  2  k  (  π      /    2  ). So in the end, your solution set is:      S    =      {                  k        π            2        ∣    k    ∈          Z        }    .

clealtAfforcewug · Accepted Answer

You are on the right track. Start with  cos  ⁡  3  x  =  cos  ⁡  xSubstituting,   cos  ⁡  3  x  =  4      cos    3    ⁡  x  −  3  cos  ⁡  x, you get  4      cos    3    ⁡  x  −  3  cos  ⁡  x  =  cos  ⁡  xThus,      cos    3    ⁡  x  −  cos  ⁡  x  =  0We can write this as  cos  ⁡  x  (      cos    2    ⁡  x  −  1  )  =  0By factoring the difference of two squares we get,  cos  ⁡  x  (  cos  ⁡  x  +  1  )  (  cos  ⁡  x  −  1  )  =  0As we expected this equation has 3 factors and hence three solutionsSet each factor equal to 0 and solve using unit circle, here n is a natural number.  cos  ⁡  x  =  0 which is satisfied for values,   x  =      ±                  (        2        n        −        1        )                  π                    2        cos  ⁡  x  +  1  =  0, thus   cos  ⁡  x  =  −  1 which is satisfied for values,   x  =  n      π    cos  ⁡  x  −  1  =  0, thus   cos  ⁡  x  =  1 which is satisfied for values   x  =  0  ,  2  n      π

How do you solve cos(3x)=cos(x)?

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