How do you find the general solution to the equation #6cos^2x-cosxsinx-sin^2x=0#?

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Answer 1 · 2018-05-20T12:00:21

Solution: #x in [0 , 2 pi),x ~~63.43^0 ,x~~ 108.43^0 # and
#x~~ 243.43^0 ,x~~ 288.43^0 #

#6 cos^2 x - cos x sin x - sin^2 x =0# or

#sin^2 x +sin x cos x = 6 cos^2 x# .Dividing by #cos^2 x# on

both sides we get, #tan^2 x +tan x = 6 # or

#tan^2 x +tan x - 6 = 0# or

#tan^2 x +3 tan x - 2 tan x - 6 = 0# or

#tan x (tan x+3 )- 2 (tan x +3) = 0# or

#(tan x+3 )(tan x- 2 ) = 0 :.# either

#(tan x+3 )=0 or (tan x- 2 ) = 0 :.#

#tan x =-3 :. x =tan^-1 (-3) ~~ -71.57^0 or 288.43^0# also

# x = (180-71.57) ~~108.43^0#

#tan x = 2:. x =tan^-1 (2) ~~ 63.43^0# also

# x = 180+63.43~~ 243.43^0#

Solution: #x in [0 , 2 pi),x ~~63.43^0 ,x~~ 108.43^0 # and

#x~~ 243.43^0 ,x~~ 288.43^0 #[Ans]