Question

Find all solutions of the equation sin^2x=2sinx+3
The answer is A+Bkpi where k is any integer and 0 12 years ago

Answer 1

I know that A is 3/2pi but I am having trouble finding B , I've tried putting 2 but apparently I'm wrong can someone help ?

Answer 2

Let y = sin(x). Put it in the given equation and you will have a quadratic equation. Solve for y
and now replace y with sin(x)

Answer 3

Got I'm going to try, thanks :)

Answer 4

*it

Answer 5

yw

Answer 6

So I've gotten y=3 and y=1

Answer 7

so if I replace I should find where sinx=3 and sinx=1 ?

Answer 8

yes. If any of the quadratic solution is invalid discard it.

Answer 9

Okay so y=1 doesn't work and so how do I find sinx=3 ?

Answer 10

sin(x) cannot be greater than 1. So discard sinx = 3.
sin(x) = 1 is good.

Answer 11

Okay so if sin(x)=1 then six(x) is pi/2 which the value 1/2 for B when I imput it into my online hw says its wrong, hmmmm I don't know what to do :/ thank you for your effort at least :)

Answer 12

No.
If sin(x) = 1 then x = pi/2 is one solution.
But remember the sine function repeats itself every 2pi.
And so there will be countless x for which sin(x) will be 1: pi/2, pi/2 + 2pi, pi/2 + 4pi, ...
So the solution for x = pi/2 + k(2pi) where k is an integer.
Compare this to A + Bkpi
A + Bkpi = pi/2 + 2kpi

So A = pi/2 and B = 2

Answer 13

Got it thank you so much !

Answer 14

you are welcome.