How do you solve #2cos^2theta+3sintheta=0#?

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Answer 1 · 2016-11-12T22:10:14

#2(1- sin^2theta) + 3sintheta= 0#

#2 - 2sin^2theta + 3sintheta = 0#

#0= 2sin^2theta - 3sintheta - 2#

#0 = 2sin^2theta - 4sintheta + sin theta - 2#

#0 = 2sintheta(sin theta - 2) + 1(sin theta -2)#

#0 = (2sintheta + 1)(sin theta - 2)#

#sintheta= -1/2 and sintheta= 2#

#theta= (7pi)/6, (11pi)/6#

Hopefully this helps!

Answer 2 · 2016-11-12T22:12:00

Please see the explanation.

Substitute #1 - sin^2(theta)# for #cos^2(theta)#:

#2(1 - sin^2(theta)) + 3sin(theta) = 0#

Use the distributive property:

#2 - 2sin^2(theta)) + 3sin(theta) = 0#

Multiply both side by -1:

#2sin^2(theta)) - 3sin(theta) - 2 = 0#

This is a quadratic where the variable is #sin(theta)#.

It looks like it will factor:

#(sin(theta) - 2)(2sin(theta) + 1) = 0

#sin(theta) = 2 and sin(theta) = -1/2#

We must discard the first root, because it is outside the range of the sine function.

Turning our attention to the second root:

#sin(theta) = -1/2#

Rotating counterclockwise from 0, the first encounter of this is at:

#theta = (7pi)/6#

The next encounter with this is at:

#theta = (11pi)/6#

Add integer rotations of #2pi# to both:

#theta = (7pi)/6 + 2npi# and #theta = (11pi)/6 + 2npi# Where n can be any integer (positive, negative, or zero)