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

Question

2229 views around the world

You can reuse this answer
Creative Commons License

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]

How do you find the general solution to the equation 6cos^2x-cosxsinx-sin^2x=06cos^2x-cosxsinx-sin^2x=0?