Question

How do I solve `5 cos 2x-6sin x-1=0` for `(0,2pi)`?

Answer 1

`x=0.41,2.73,(3pi)/2`

Explanation:

`"using the "color(blue)"trigonometric identity"`

`•color(white)(x)cos2x=1-2sin^2x`

`rArr5(1-2sin^2x)-6sinx-1=0`

`rArr5-10sin^2x-6sinx-1=0`

`rArr-10sin^2x-6sinx+4=0`

`rArr-2(5sin^2x+3sinx-2)=0`

`"we have a quadratic in sine"`

`rArr-2(5sinx-2)(sinx+1)=0`

`rArr5sinx-2=0" or "sinx+1=0`

`•color(white)(x)sinx=2/5`

`rArrx=0.41" or "x=(pi-0.41)=2.73to("2 dec. places")`

`•color(white)(x)sinx=-1rArrx=(3pi)/2`

`rArrx=0.41,2.73,(3pi)/2;x in(0,2pi)`