PLEASE help me with this problem?

Question

Solve for x, if 0 < x < 2pie. Write your answers in exact values only.

sinx + 2sinx cosx = 0

Answer 1

sinx + 2sinx cosx = 0 <=> sinx (1 + 2cosx) = 0
<=> sin x = 0 or 1 + 2cosx = 0 <=>
sinx = 0 <=> x = k*pi ; k = ..,-1,0,1,...
1 + 2cosx = 0 <=> cosx = -1/2 <=> x = 3/4 pi + kpi ; k = ..,-1,0,1,...

anything else yoiu want to know ?

Answer 2

Factor out a "sin x":

sin x (1 + 2 cos x) = 0

Two options:

sin x = 0, on 0 < x < 2pi

sin x is opposite / hypotenuse. Since hypotenuse cannot be zero, the two places where "opposite" (y-coordinate) is zero are
x = 0 and x = pi

Second option:
1 + 2 cos x = 0
cos x = (-1/2)
cos x is adjacent / hypotenuse. hypotenuse = 2
and adjacent = -1. This is the 30-60-90 triangle and since the adjacent (the x-coordinate) is -1, the reference angle is 60 degrees or (pi / 3). But since cos x < 0, the two quadrants where this is defined are the 2nd and 3rd quadrants.
So x = pi - (pi / 3) or pi + (pi / 3)
=2pi/3 or 4pi/3

x = 0, 2pi/3, pi , 4pi/3

Answer 3

sinx(1+2cosx) = 0
so, either sinx = 0 ...(1)
or, cosx = - 1/2 ........(2)
and 0 < x < 2pie
so from (1) x= pie
n from (2) x = 2pie/3 + 2npie n 4pie/3 +2npie
hence we'll get 7 answers

Answer 4

x = 2.0943951 (or 2/3*pie), 3.14159 (or pie), and 4.1887902 (or 4/3*pie),

so technically 3 answers if 0 and 2*pie aren't allowed as your equation shows (however if those were meant to be equal to or greater, then zero and 2*pie also work, for a total of five).

sin x + 2 sin(x)*cos(x) = 0
sinx = -2sin(x)*cos(x)
-1/2 = cosx
x = 2.0943951 and 4.1887902
or when sinx = 0 which is at 0, pie and 2*pie.

You can visually see these answers if you sketch the functions and see where sine crosses zero and where cosine is at -0.5

final answer

Answer 5

Sin(x) + 2Sin(x).Cos(x) = 0
Sin(x)*[1 + 2Cos(x)] = 0
Sin(x)=0 and/or 1 + 2Cos(x) + 0
If Sin(x) = 0, then x = 0 or x = pie or x = 2pie.
If 1 + 2Cos(x) = 0, then 2Cos(x) = -1. So, Cos(x) = -1/2. It means that x = 2/3 pie or x = 4/3pie
Totally there are five answers: x=0, x=2/3pie, x=pie, x=4/3pie, x=2pie.

Answer 6

Come on !

Hint : sinx + 2sinx cosx = 0 = sinx(1 + 2cosx)

Easy now, no ?

Answer 7

sinx(1+cosx)=0
sinx=0 so x=0
1+2cosx=0 or cosx=-1/2
so x=4pi/3
so x can be 0 or 4pi/3

Answer 8

i had to do this in my last module of maths...periodic functions i think it was. have you tried graphing it? after you do that write down where it intercepts the x axis.

Answer 9

1 would work i think

Answer 10

wow...good luck with that....I got an A in algebra. I just got a level 2 though, so thanks a bunch!