How do you solve #sin^2x+3sinx-4=0# for #0<=x<=2pi#?
2 Answers
Nov 28, 2016
Explanation:
this is a quadratic eqn. in
factorise if possible.
but
for
Nov 28, 2016
Explanation:
Solve this quadratic equation for sin x:
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
The value sin x = -4 is rejected as < - 1.
Solve sin x = 1
trig unit circle -->
sin x = 1 --> arc
Answers for
x = pi/2
Check: