Is this possible to factor? #3sin^2θ-cos^2θ-1#

Question

#3sin^2θ-cos^2θ-1#

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Answer 1 · 2018-05-17T03:45:10

Not in its current form but a substitution makes it factorable as the difference of two squares.

Given #3sin^2(theta)-cos^2(theta)-1#

Substitute #cos^2(theta) = 1- sin^2(theta)#

#3sin^2(theta)-(1-sin^2(theta))-1#

Distribute the minus:

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

Combine like terms:

#4sin^2(theta)-2#

The above factors as the difference of two squares:

#(2sin(theta)-sqrt2)(2sin(theta)+sqrt2)#