please solve the trig equation

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February 16th 2011, 12:18 PM
kassums17

please solve the trig equation

sin2xcosx+cos2xsinx=1
February 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

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

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

$2\sin(x) (1-\sin^2(x)) + \sin(x) - 2\sin^3(x) = 1 \implies 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x) = 1$
February 16th 2011, 12:32 PM
pickslides

$\displaystyle \sin 2x \cos x+\cos 2x \sin x = \sin (2x+x)$
February 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

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

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

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

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!!!
February 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 $\displaystyle 2X - 2X^3 + X - 2X^3 = 1$ , with $\displaystyle X = \sin{x}$ .

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

Quote:

Originally Posted by Prove It

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

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

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
February 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.
February 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
February 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.
February 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?
February 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 $\displaystyle \sin{(A + B)} \equiv \sin{A}\cos{B} + \cos{A}\sin{B}$ .

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

Quote:

Originally Posted by Prove It

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

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

Ok then, i use that to get

sin(2x+x) = 1

what can i do from that?
February 16th 2011, 06:06 PM
Prove It

Surely you can solve $\displaystyle \sin{3x} = 1$ ...
February 16th 2011, 06:16 PM
TacticalPro

Quote:

Originally Posted by Prove It

Surely you can solve $\displaystyle \sin{3x} = 1$ ...

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

so anyway it should be:

sin2x+sinx=1

sin3x=1

sinx=1/3
February 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!

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

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