sin2xcosx+cos2xsinx=1

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- Feb 16th 2011, 12:18 PMkassums17please solve the trig equation
sin2xcosx+cos2xsinx=1

- Feb 16th 2011, 12:30 PMe^(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 PMpickslides
$\displaystyle \displaystyle \sin 2x \cos x+\cos 2x \sin x = \sin (2x+x)$

- Feb 16th 2011, 04:56 PMTacticalPro
- Feb 16th 2011, 05:03 PMProve It
- Feb 16th 2011, 05:13 PMTacticalPro
- Feb 16th 2011, 05:32 PMProve 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 PMTacticalPro
- Feb 16th 2011, 05:46 PMmr fantastic
- Feb 16th 2011, 05:51 PMTacticalPro
- Feb 16th 2011, 05:53 PMProve It
- Feb 16th 2011, 06:04 PMTacticalPro
- Feb 16th 2011, 06:06 PMProve It
Surely you can solve $\displaystyle \displaystyle \sin{3x} = 1$...

- Feb 16th 2011, 06:16 PMTacticalPro
- Feb 16th 2011, 06:19 PMProve It