Question

How do you solve `sin^2x+3sinx-4=0` for `0<=x<=2pi`?

Answer 1

`x=3/2pi`

Explanation:

this is a quadratic eqn. in `" "sinx`

factorise if possible.

`(sinx+4)(sinx-1)=0`

`=>sinx+4=0" "`or`" "sinx-1=0`

`sinx+4=0" "=> sinx=-4`

but `-1 < sinx <1` for `x inRR`

`:.`no real solutions for `" "sinx=-4`

`sinx-1=0" "=>sinx=1`

for `0 < x <2pi`

`x=pi/2`

Answer 2

`pi/2`

Explanation:

Solve this quadratic equation for sin x:
`sin^2 x + 3sin x - 4 = 0`
We have a = 1, b = 3, and c = -4.
Since a + b + c = 0, use shortcut.
The 2 real roots are sin x = 1 and `sin x = c/a = -4/1 = -4`
The value sin x = -4 is rejected as < - 1.
Solve sin x = 1
trig unit circle -->
sin x = 1 --> arc `x = pi/2`
Answers for `(0,2pi)`:
x = pi/2
Check:
`x = pi/2 -> 3sin (pi/2) = 3 -> sin^2 x = 1 -> 3 + 1 - 4 = 0`. OK