Question

How do you solve `3cscA-2sinA-5=0`?

Answer 1

`A=kpi+(-1)^k(pi/6),kinZ`

Explanation:

`3cscA-2sinA-5=0`
`rArr3/sinA-2sinA-5=0`
`rArr3-2sin^2A-5sinA=0`
`rArr2sin^2A+5sinAcolor(red)(-3)=0`
`rArr2sin^2A+6sinA-sinA-3=0`
`rArr2sinA(sinA+3)-1(sinA+3)=0`
`rArr(sinA+3)(2sinA-1)=0`
`rArrsinA=-3!in[-1,1],sinA=1/2in[-1,1]`
`rArrsinA=sin(pi/6)`
`rArrA=kpi+(-1)^k(pi/6),kinZ`
`rArrA=kpi+(-1)^k(pi/6),kinZ`

Answer 2

`A=(npi)/2+-pi/3, ninZZ`
`color(white)(A)=n90^circ+-60^circ, ninZZ`

Explanation:

First we multiply everything by `sinA` since `cscA=1/sinA` and sinA/sinA=1#

`sinA(3cscA-2sinA-5)=sinA(0)`

`3-2sin^2A-5sinA=0`

Substitute `x=sinA`

`2x^2+5x-3=0`

`x=(-b+-sqrt(b^2-4ac))/2`

`color(white)(x)=(-5+-sqrt(5^2-4(2*-3)))/4`

`color(white)(x)=(-5+-sqrt(25+24))/4`

`color(white)(x)=(-5+-sqrt(49))/4`

`color(white)(x)=(-5+-7)/4`

`color(white)(x)=(-5-7)/4 or (-5+7)/4`

`color(white)(x)=-12/4 or 2/4`

`color(white)(x)=-3 or 1/2`

However, `-1lesinAle1` so we must take `1/2`

`sinA=1/2`

`A=arcsin(1/2)=pi/6-=30^circ, A=(5pi)/6-=150^circ`

`A=(npi)/2+-pi/3, ninZZ`
`color(white)(A)=n90^circ+-60^circ, ninZZ`