Question

How do you solve the following equation `6cos^2x = 19sinx + 16` in the interval [0, 2pi]?

Answer 1

`x=(180^o)+41^o49'=221^o49' in QIII`
`Q IV, x=(360^o)-(41^o49')=318^o11'`,

Explanation:

`6cos^2x=19sinx+16`
`rArr 6(1-sin^2x)=19sinx+16`
`rArr 6sin^2x+19sinx+10=0`
`rArr 6sin^2x+15sinx+4sinx+10=0`
`rArr 3sinx(2sinx+5)+2(2sinx+5)=0`
`rArr (2sinx+5)(3sinx+2)=0`
`rArr sinx =-5/2,` impossible , as, `sinx in [-1,1], sinx=-2/3`
`sinx =-2/3 ~= -0.6667 = -sin(41^o49')`

Since `sinx<0` and, `x in [0,2pi]`, x lies in Quadrant III, IV.

Accordingly, `x=(180^o)+41^o49'=221^o49' in QIII`
For `Q IV, x=(360^o)-(41^o49')=318^o11'`,