Solve sin(x)(sinx+1)=0

Question

Solve sin(x)(sinx+1)=0

 

a)+/- πn, x=π/2 +/-2πn

b) π/n+/- 2πn, x=3π/2+2πn

c) x=+/- πn

d) +/-πn, x=3π/2 +/- 2πn

 

 

I will attempt to try this... but i dont understand... 

 

sinx=0 π, and 2π

 

and sinx=-1 at 3π/2, 

 

 

Uhm....I'm lost?

can you show the steps?

Answer 1

Greetings! Lets solve this shall we ?

Then, we must find all values where the function sinx(sinx + 1) equal 0 on the unit circle. We can find these values algebraically such that

sinx(sinx + 1) = 0.............Using the Zero-Product Property of Equality as you did such that

sinx = 0 and sinx + 1 = 0.................We solve each one for x such that

x = sin^-1(0) = 0, π, 2π, 3π, ..., nπ, where n = 1, 2, ..., k means sinx = 0 at 0, pi, 2pi, 3pi, etc per half-revolution.

Then, we have sinx + 1 = 0 such that

sinx + 1 = 0.............Subtract 1 to both sides such that

sinx = -1............Then sine inverse both sides such that

x = sin^-1(-1) = 3π/2......Begins at 3π/2 then sinx = -1 and for every 1 whole revolution after such that

x = 3π/2 + 2πn

I hope this helped!