sin(x+Ï€/4) = 2 cos(x-Ï€/4) Solve for x

We are asked
to solve sin(x+pi/4)=2(cos(x-pi/4)) for x.

Often, to solve equations
involving trigonometric functions, it helps to try to transform the equation to one that
involves only one function. Usually you will use identities to achieve that goal.

Here we can use the sum identities for sine and cosine. Recall that:

sin(A+B)=sinAcosB + sinB
cosA
cos(A-B)=cosAcosB+sinAsinB

Using these we
rewrite the given equation as:

sinxcos(pi/4)+cosxsin(pi/4)=2(cosxcos(pi/4)+sinxsin(pi/4))

Noting that sin(pi/4)=cos(pi/4)= `sqrt(2)/2` , we substitute into
the equation to get:

`sqrt(2)/2 sinx+sqrt(2)/2
cosx=2(sqrt(2)/2 cosx + sqrt(2)/2 sinx)`

Using the
distributive property and collecting like terms we get sinx=-cosx.

(1) We can
divide both sides by cosx to get tanx=-1. The arctan (or inverse tangent) of -1 is -pi/4. Since
tangent has a period of pi, the full answer is:

`x=-pi/4 +
npi, n in ZZ` (n is an integer.)

(2) We could recognize
that sinx=cosx implies x is a multiple of pi/4. The sine and cosine have opposite sign in the
2nd and 4th quadrants giving the same answer as above.

You might know, or be
able to show, that cos(x-pi/4)=sin(x+pi/4). Then the original equation becomes
sin(x+pi/4)=2(sin(x+pi/4)) or sin(x+pi/4)=0. Then since sinA=0 implies A=0 we have x+pi/4=0 or
x=-pi/4 as above.

Last, you might graph the left and right sides of the
equation and look for the intersection.

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