How do you solve #4cos^2x-5sinxcosx-6=0#?

1 Answer
Nghi N.
Dec 3, 2015

solve trig equation.

Explanation:

#4cos^2 x - 5sin x.cos x - 6 = 0#
Try to transform the equation into a linear function in sin 2x and
cos 2x. Replace #4cos^2 x = 2(1 + cos 2x)# and
5sin x.cos x by #5/2sin 2x.#
#2 + 2cos 2x - (5/2)sin 2x - 6 = 0#
#2cos2x - 5/2sin 2x - 4 = 0#
#cos 2x - (5/4)sin 2x = 2#
Call#tan a = sin a/(cos a) = 5/4# --> #a = 51^@34# --> #cos a = 0.62#
#cos 2x.cos a - sin a.sin 2x = 2cos a#
#cos (2x + 51.34) = 2(0.62) = 1.25 > 1.#
This trig equation is insolvable since any cos must be < 1.
The number 6 should be 5 or less in order to make the equation solvable.

Related questions
Impact of this question
2504 views around the world
You can reuse this answer
Creative Commons License