# please solve the trig equation

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• Feb 16th 2011, 12:18 PM
kassums17
sin2xcosx+cos2xsinx=1
• Feb 16th 2011, 12:30 PM
e^(i*pi)
Use the double angle identities, I have used the sin form of cos(2x) because I want to eliminate cos

$\displaystyle 2\sin(x)\cos(x)\cos(x) + (1- 2\sin^2(x))\sin(x) = 1 \implies 2\sin(x) \cos^2(x) + \sin(x) - 2\sin^3(x) = 1$

Of course $\displaystyle \cos^2(x) = 1-\sin^2(x)$ and hence we have

$\displaystyle 2\sin(x) (1-\sin^2(x)) + \sin(x) - 2\sin^3(x) = 1 \implies 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x) = 1$
• Feb 16th 2011, 12:32 PM
pickslides
$\displaystyle \displaystyle \sin 2x \cos x+\cos 2x \sin x = \sin (2x+x)$
• Feb 16th 2011, 04:56 PM
TacticalPro
Quote:

Originally Posted by e^(i*pi)
Use the double angle identities, I have used the sin form of cos(2x) because I want to eliminate cos

$\displaystyle 2\sin(x)\cos(x)\cos(x) + (1- 2\sin^2(x))\sin(x) = 1 \implies 2\sin(x) \cos^2(x) + \sin(x) - 2\sin^3(x) = 1$

Of course $\displaystyle \cos^2(x) = 1-\sin^2(x)$ and hence we have

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

I followed your math up to where you stopped. What do you do now? At first I thought the two cubed sins will cancel each other out but they're both -2sin^3, so what do you do next? please respond. thanks!!!
• Feb 16th 2011, 05:03 PM
Prove It
Quote:

Originally Posted by TacticalPro
I followed your math up to where you stopped. What do you do now? At first I thought the two cubed sins will cancel each other out but they're both -2sin^3, so what do you do next? please respond. thanks!!!

Rewrite the equation as $\displaystyle \displaystyle 2X - 2X^3 + X - 2X^3 = 1$, with $\displaystyle \displaystyle X = \sin{x}$.

Solve for $\displaystyle \displaystyle X$, then use this to solve for $\displaystyle \displaystyle {x}$.
• Feb 16th 2011, 05:13 PM
TacticalPro
Quote:

Originally Posted by Prove It
Rewrite the equation as $\displaystyle \displaystyle 2X - 2X^3 + X - 2X^3 = 1$, with $\displaystyle \displaystyle X = \sin{x}$.

Solve for $\displaystyle \displaystyle X$, then use this to solve for $\displaystyle \displaystyle {x}$.

Ok i replaced sinx with x and simplified it to (-4x^3) + (3x) - 1

what do I do next? I tried factoring out an x but got no where
• Feb 16th 2011, 05:32 PM
Prove It
Use the Factor Theorem to find a linear factor and use long division to factorise this cubic to its linear and quadratic factors. Factorise the quadratic if possible. Then solve using Null Factor Law.
• Feb 16th 2011, 05:43 PM
TacticalPro
Quote:

Originally Posted by Prove It
Use the Factor Theorem to find a linear factor and use long division to factorise this cubic to its linear and quadratic factors. Factorise the quadratic if possible. Then solve using Null Factor Law.

I'm sorry but I have no idea what you just said. I looked up factor theorem and had no luck deciphering. could you please explain further? thanks
• Feb 16th 2011, 05:46 PM
mr fantastic
Quote:

Originally Posted by TacticalPro
I'm sorry but I have no idea what you just said. I looked up factor theorem and had no luck deciphering. could you please explain further? thanks

Post #2 tells you exactly what to do. Why would you want to do it in a much more complicated way. When you want to kill a fly, use a fly swatter. Not an elephant gun.
• Feb 16th 2011, 05:51 PM
TacticalPro
Quote:

Originally Posted by mr fantastic
Post #2 tells you exactly what to do. Why would you want to do it in a much more complicated way. When you want to kill a fly, use a fly swatter. Not an elephant gun.

I looked at post 2 but don't understand it? all he did was change the 1 to a sin(2x+x)?? what identity supports that? what is he implying?
• Feb 16th 2011, 05:53 PM
Prove It
Quote:

Originally Posted by TacticalPro
I looked at post 2 but don't understand it? all he did was change the 1 to a sin(2x+x)?? what identity supports that? what is he implying?

The identity they used is $\displaystyle \displaystyle \sin{(A + B)} \equiv \sin{A}\cos{B} + \cos{A}\sin{B}$.

Here $\displaystyle \displaystyle A = 2x$ and $\displaystyle \displaystyle B = x$.
• Feb 16th 2011, 06:04 PM
TacticalPro
Quote:

Originally Posted by Prove It
The identity they used is $\displaystyle \displaystyle \sin{(A + B)} \equiv \sin{A}\cos{B} + \cos{A}\sin{B}$.

Here $\displaystyle \displaystyle A = 2x$ and $\displaystyle \displaystyle B = x$.

Ok then, i use that to get

sin(2x+x) = 1

what can i do from that?
• Feb 16th 2011, 06:06 PM
Prove It
Surely you can solve $\displaystyle \displaystyle \sin{3x} = 1$...
• Feb 16th 2011, 06:16 PM
TacticalPro
Quote:

Originally Posted by Prove It
Surely you can solve $\displaystyle \displaystyle \sin{3x} = 1$...

shoot me.... i cramped, thought it was sin^(2x+x)=1

so anyway it should be:

sin2x+sinx=1

sin3x=1

sinx=1/3
• Feb 16th 2011, 06:19 PM
Prove It
Quote:

Originally Posted by TacticalPro
shoot me.... i cramped, thought it was sin^(2x+x)=1

so anyway it should be:

sin2x+sinx=1

sin3x=1

sinx=1/3

Absolutely not!

$\displaystyle \displaystyle \sin{3x} = 1 \implies 3x = \arcsin{1} + 2\pi n$ with $\displaystyle \displaystyle n \in \mathbf{Z}$...
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