1. ## please solve the trig equation

sin2xcosx+cos2xsinx=1

2. 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

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

3. $\displaystyle \sin 2x \cos x+\cos 2x \sin x = \sin (2x+x)$

4. 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!!!

5. 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 $\displaystyle {x}$.

6. 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

7. 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.

8. 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

9. 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.

10. 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?

11. 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 $\displaystyle B = x$.

12. 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?

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

14. 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

15. 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 $\displaystyle n \in \mathbf{Z}$...

Page 1 of 2 12 Last