Question

Solving Trigonometric Equation. Rewrite the equation 3cscx - sinx = 2 in terms of sine. Then, solve algebraically for 0 < x < 2pi?

Sorry for asking too many questions.

Answer 1

`x_(1_1)=pi/2 or x_(1_2)=-7/4pi`

Explanation:

`3cscx - sinx = 2|*sin(x)`
`3-sin(x)^2=2sin(x)|+sin(x)^2|-3`
`0=sin(x)^2+2sin(x)+1-1-3`
`0=(sin(x)+1)^2-4|+4`
`4=(sin(x)+1)^2|sqrt()`
`+-2=sin(x)+1|-1`
`-1+-2=sin(x)`

`arcsin(-1+-2)=x`
`x_1=arcsin(1)+2pin, n∈ZZ`
`x_1=pi/2+2pin, n∈ZZ`

`x_2=arcsin(-3)+2pin`
`arcsin(-3) " can't be expressed with real numbers."`

`arcsin(-3) and x_2 ∈CC`

`"For " 0 < x < 2pi:`
`x_(1_1)=pi/2+2pi*0=pi/2 or x_(1_2)=pi/2-1*2pi=-7/4pi`

Answer 2

`pi/2`

Explanation:

3csc x - sin x = 2
`3/(sin x) - sin x = 2`
`3 - sin^2 x = 2sin x`
Solve this quadratic equation for sin x
`sin^2 x + 2sin x - 3 = 0`
Since a + b + c = 0, use shortcut. There are 2 real roots:
sin x = 1 , and `sin x = c/a = - 3` (rejected)
Answer -->
sin x = 1 --> `x = pi/2`