Question 77922
{{{sin(4x) + sin x = 0}}} Start with the given expression


{{{sin(2*(2x)) + sin x = 0}}} Rewrite {{{sin(4x)}}} into {{{sin(2*(2x))}}}


{{{2sin(2x)cos(2x) + sin x = 0}}} Use the identity: {{{sin2x=2sinxcosx}}}


{{{2(2sin(x)cos(x))cos(2x) + sin x = 0}}} Use the identity: {{{sin2x=2sinxcosx}}} again


{{{(4sin(x)cos(x))(2(cos(x))^2-1) + sin x = 0}}} Use the identity: {{{cos2x=2(cos(x))^2-1}}}



{{{8sin(x)(cos(x))^3-4sin(x)cos(x) + sin x = 0}}} Distribute {{{4sin(x)cos(x)}}}


{{{sin(x)(8(cos(x))^3-4cos(x) + 1) = 0}}} Factor out a sin(x)


Since we know the value of x for {{{sin(x)=0}}} (the solution is x=0) we can ignore the sin(x) and try to solve the expression in the parenthesis


{{{8(cos(x))^3-4cos(x) + 1 = 0}}} So lets focus on the terms in the parenthesis

Let {{{y=cos(x)}}}

So we get

{{{8y^3-4y+1=0}}}


{{{8y^3-2y-2y+1=0}}} Rewrite -4y into -2y-2y. This will allow us to factor


{{{2y(4y^2-1)-(2y-1)=0}}} Group like terms and factor out the GCF


{{{2y(2y+1)(2y-1)-(2y-1)=0}}} Factor {{{4y^2-1}}} into {{{(2y+1)(2y-1)}}} using difference of squares



{{{(2y(2y+1)-1)(2y-1)=0}}} Combine like terms (note: the common term is {{{2y-1}}})


Now set each factor equal to zero. Lets start with {{{2y-1}}}

{{{2y-1=0}}}

{{{2y=1}}}

{{{y=1/2}}}

Now let {{{cos(x)=1/2}}} and solve for x  note: I'm using radians

{{{x=1.04719+pi*n}}} and {{{x=-1.04719+pi*n}}}  Since our interval is [0,2pi] we must ignore the negative answer. So of these two answers, {{{x=1.04719+pi*n}}} is the only solution.





Now let {{{2y(2y+1)-1=0}}}

{{{4y^2+2y-1=0}}} Distribute the 2y

Use the quadratic formula to solve for y

*[invoke quadratic "y", 4, 2, -1 ]

So if we replace y with {{{cos(x)}}} we get these solutions


{{{cos(x)=0.309017}}} or {{{cos(x)=-0.809017}}}


Take the arccosine of both sides (for the solution {{{cos(x)=0.309017}}}) to solve for x


{{{x=1.25664+pi*n}}} or {{{x=-1.25664+pi*n}}} Here are 2 more possible solutions. Since our interval is [0,2pi] we must ignore the negative answer. So of these two answers, {{{x=1.25664+pi*n}}} is the only solution.


Now lets use the other answer of {{{cos(x)=-0.809017}}}


{{{cos(x)=-0.809017}}}

Take the arccosine of both sides

{{{x=2.51327+pi*n}}} or {{{x=-2.51327+pi*n}}} Since our interval is [0,2pi] we must ignore the negative answer. So of these two answers, {{{x=2.51327+pi*n}}} is the only solution.

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So after all of that, we find that our solutions are

{{{x=0}}} (from sin(x)=0) or {{{x=1.04719+pi*n}}} or {{{x=1.25664+pi*n}}} or {{{x=2.51327+pi*n}}}

As always, we can check our work by using a calculator.

This is a lot to take in, so feel free to ask me further about any of this.