How do I solve this problem?

Solve the equation by first using a sum to product formula:
#sin4theta-sin2theta=cos3theta#

Solve the equation by first using a sum to product formula:
#sin4theta-sin2theta=cos3theta#

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Noah G Share
Feb 9, 2018

Answer:

I got #x= pi/6 + pi/2n# and #x = pi/2 + pin#

Explanation:

Let's start by rewriting:

#sin(2theta + 2theta) - sin(2theta) = cos(2theta + theta)#

#sin(2theta)cos(2theta) + sin(2theta)cos(2theta) - sin(2theta) = cos(2theta)costheta - sin(2theta)sintheta#

#2sin(2theta)cos(2theta) - sin(2theta) = cos(2theta)costheta - sin(2theta)sintheta#

#sin(2theta)(2cos2theta - 1) = cos(2theta)costheta - 2sinthetacosthetasintheta#

#sin2theta(2cos2theta - 1) = costheta(cos2theta - 2sin^2theta)#

Now recall that #cos(2theta) = 1 - 2sin^2theta#.

#((2sinthetacostheta)(2cos2theta - 1))/costheta = 1 - 4sin^2theta#

#2sintheta(2(1 - 2sin^2theta) - 1) = 1 - 4sin^2theta#

#2sintheta(2 - 4sin^2theta - 1) = 1 - 4sin^2theta#

#4sintheta - 8sin^3theta - 2sintheta = 1 -4sin^2theta#

#0= 8sin^3theta - 4sin^2theta - 2sintheta + 1#

Now let #y = sin theta#.

#0 = 8y^3 - 4y^2 - 2y + 1#

#0 = 4y^2(2y - 1) -(2y - 1)#

#0 = (4y^2 - 1)(2y - 1)#

#y^2 = 1/4 or y = 1/2#

#y= +-1/2#

#sintheta = +-1/2#

#theta = pi/6 + pi/2n#

However, we must account for when we cancelled the cosine above. This can be a dangerous game, because you can lose solutions. However, if you recognize it, you can prevent that from happening.

Since the equation at this point is a product, the equation will hold true whenever #costheta = 0# (rendering the entire equation #0#).

Thus, #theta = pi/2 + pin#. The graph of the functions #y_1 = sin(4x) - sin(2x)# and #y_2 = cos(3x)# confirm that the solutions we found are indeed correct.

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Hopefully this helps!

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