How do you solve the trig function, sin squared x + sin x = 0?

Question

The domain is [0,2pi]. I tried using the trig identities, but I still couldn't solve it.

Answer 1

(Sin(x))²+Sin(x) ==0

let t== Sin(x) be.

the equation becomes :

t²+t==0 iff

t(t+1) ==0

the solutions are:

t==0 and t==-1

but t== Sin(x) then:

Sin(x)==0 <===> x==0 and TT

Sin(x) ==-1 <===> x== 3/2 TT

so the solutions are:

x1==0, x2== TT, x3== (3/2) TT

proof:

for x==0

Sin²(x)==0 and Sin(x)== 0 so

(Sin(x))²+Sin(x) ==0 is true

for x== TT

Sin(TT)²== 0 and Sin(TT)==0 ok

for x==(3/2)TT

Sin((3/2)TT)²== (-1)²==1

Sin((3/2)TT)==-1

(Sin((3/2)TT))²+Sin((3/2)TT) ==1-1 ==0 ok

Answer 2

You set up a variable.  Say sin x = y.  Now we substitute into the equation to get y² + y=0.  Then we solve the quadratic to get y={0,-1}.  Then we substitute sin x back for y to get sin x = {0, -1}.  We solve for the individual values of 0 and -1 to get x={0, pi, 3pi/2}.

Hope it helps.  =D