1. ## trigo identities

solve the eqns for 0<=x<=180

1. sin3x+sinx=0
(i don't get my friend's working):
2 sin2xcosx=0
2(2sinxcosx)(cosx)=0
(4sinxcosx)(cosx)=0
4 sinx=0
sinx=0
x= 0, 180

what I don't get is how the (4 sin x cos x) (cos x) = 0 became just 4 sin x = 0.
she said she canceled it and gave no further explanation. i don't get it, because it's multiply, how could she have canceled the cos x-es out.

2. sin5x-sinx=0
i got up until:
2 cos 3x sin 2x = 0
2 (4cos*3x-3cosx) (2 sinx cosx) = 0

2. Originally Posted by colloquial
solve the eqns for 0<=x<=180

1. sin3x+sinx=0
(i don't get my friend's working):
2 sin2xcosx=0
2(2sinxcosx)(cosx)=0
(4sinxcosx)(cosx)=0
4 sinx=0
sinx=0
x= 0, 180

what I don't get is how the (4 sin x cos x) (cos x) = 0 became just 4 sin x = 0.
she said she canceled it and gave no further explanation. i don't get it, because it's multiply, how could she have canceled the cos x-es out.
I don't know what's going on either.

Is this
$sin^3(x) + sin(x) = 0$
or
$sin(3x) + sin(x) = 0$?

Based on your friend's answer I'm guessing it's the first one. So factor the common sin(x):
$sin(x)(sin^2(x) + 1) = 0$

Thus either $sin(x) = 0$ ==> $x = 0^o, 180^o$
or $sin^2(x) + 1 = 0$ which is valid for no values of x.

So the solution is $0^o, 180^o$.

-Dan

3. it's the second one.

eta:
this is what I got -

(4 sin x cos x) (cos x) = 0
cos^2 x . 4 sin x = 0

i.
cos^2x = 0
cos x = 0
x = 90

ii.
4 sin x = 0
sin x = 0
x = 0, 180

therefore, x = 0, 90, 180

is this correct?

4. anyone can help me with #2? please?
thanks!

5. Originally Posted by colloquial
it's the second one.

eta:
this is what I got -

(4 sin x cos x) (cos x) = 0
Where are you getting this line from?

$sin(3x) + sin(x) = 0$

First:
$sin(3x) = sin(2x + x) = sin(2x)cos(x) + sin(x)cos(2x)$

$= 2sin(x)cos(x) \cdot cos(x) + sin(x)(1 - 2sin^2(x))$

$= 2sin(x)cos^2(x) + sin(x) - 2sin^3(x)$

$= 2sin(x)(1 - sin^2(x)) + sin(x) - 2sin^3(x)$

$= 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x)$

$= 3sin(x) - 4sin^3(x)$

$[3sin(x) - 4sin^3(x)] + sin(x) = 0$

$-4sin^3(x) + 4sin(x) = 0$

$sin^3(x) - sin(x) = 0$
which looks disturbingly like the other problem... Please check my Math on this!

-Dan

6. Hello, colloquial!

Solve for $0^o \leq x \leq 180^o$

$1)\;\;\sin3x + \sin x\:=\:0$
Sum-to-Product identity: . $\sin A + \sin B \:=\:2\!\cdot\!\sin\!\left(\frac{A+B}{2}\right)\!\ cdot\!\cos\!\left(\frac{A-B}{2}\right)$

The equation becomes: . $2\cdot\sin(2x)\cdot\cos(x)\:=\:0$

We have two equations to solve:

. . $\sin(2x)\,=\,0\quad\Rightarrow\quad 2x \:=\:0^o,\:180^o,\:360^o\quad\Rightarrow\quad x \:=\:0^o,\:90^o,\:180^o$

. . $\cos x \:=\:0\quad\Rightarrow\quad x \,=\,90^o$

Solutions: . $\boxed{x \:=\:0^o,\:90^o,\:180^o}$

$2)\;\;\sin5x - \sin x \:=\:0$
Another identity: . $\sin A - \sin B \:=\:2\!\cdot\!\cos\!\left(\frac{A+B}{2}\right)\!\ cdot\!\sin\!\left(\frac{A-B}{2}\right)$

The equation becomes: . $2\cdot\cos(3x)\cdot\sin(2x) \:=\:0$

We have two equations to solve:

. . $\cos3x \,=\,0\quad\Rightarrow\quad 3x\:=\:90^o,\:270^o,\:450^o\quad\Rightarrow\quad x \:=\:30^o,\:90^o,\:150^o$

. . $\sin2x \,=\,0\quad\Rightarrow\quad 2x\:=\:0^o,\:180^o,\:360^o\quad\Rightarrow\quad x \:=\:0^o,\:90^o,\:180^o$

Solution: . $\boxed{x\;=\;0^o,\:30^o,\:90^o,\:150^o,\:180^o}$

7. Originally Posted by Soroban
Sum-to-Product identity: . $\sin A + \sin B \:=\:2\!\cdot\!\sin\!\left(\frac{A+B}{2}\right)\!\ cdot\!\cos\!\left(\frac{A-B}{2}\right)$
I always forget that one...

-Dan