SOLUTION: Determine the solutions to sin^2 θ = -sin θ where -π≤x<π

Algebra ->  Trigonometry-basics -> SOLUTION: Determine the solutions to sin^2 θ = -sin θ where -π≤x<π      Log On


   

Question 1193294: Determine the solutions to sin^2 θ = -sin θ where -π≤x<π
Found 3 solutions by ikleyn, greenestamps, Solver92311:
Answer by ikleyn(52816) About Me  (Show Source):
You can put this solution on YOUR website!
.
Determine the solutions to sin^2 θ = -sin θ where -π≤x<π
~~~~~~~~~~~~~ 

One solution is obvious: it is sin(θ) = 0,  which gives two answers  θ = 0  and 
θ = -pi,  belonging to the given interval.


To find another solution with sin(θ) =/= 0,  divide the original equation by sin(θ), both sides.


You will get then  sin(θ) = -1, which has the solution  θ = -pi%2F2  belonging to the given interval.


Thus there are three answers in the given interval:  θ = 0,  θ = -pi%2F2  and  θ = -pi.    ANSWER

Solved.



Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


%28sin%28x%29%29%5E2=-sin%28x%29

This is a quadratic equation with "sin(x)" as the "variable". Solve like any quadratic equation: get everything on one side set equal to 0 and factor.

%28sin%28x%29%29%5E2%2Bsin%28x%29=0
sin%28x%29%28sin%28x%29%2B1%29=0

sin(x)=0 OR sin(x)=-1

On the specified interval, sin(x)=0 at x=-pi and x=0; sinx=-1 at x=-pi/2.

ANSWERS: x=-pi, x=-pi/2, and x=0

Answer by Solver92311(821) About Me  (Show Source):
You can put this solution on YOUR website!

Since and are two entirely different things, is meaningless with respect to the given equation. Therefore you must need the solution for all real values of

[1]


[2]

Since ,
is a partial solution.

Furthermore, dividing eq. [2] by gives

[3]

Since ,
is also part of the solution set.

In sum, the solution set is

John

My calculator said it, I believe it, that settles it

From
I > Ø