Solve #4cos^2x-5sinxcosx-6=0#?

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Answer 1 · 2017-11-23T15:18:21

No solution.

Dividing each term in #4cos^2x-5sinxcosx-6=0# by #cos^2x#

and we get #4-5tanx-6sec^2x=0#

or #4-5tanx-6(1+tan^2x)=0#

or #4-5tanx-6-6tan^2x=0#

or #6tan^2x+5tanx+2=0#

Now as the determinant is #5^2-4xx6xx2=-23<0#, there are no real roots.

Hence, no solution.