4sinx-2cos(2x)=1

Please help, how to solve this equation

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- Aug 29th 2012, 12:48 PMTeloTrigonometric Equation Problem
4sinx-2cos(2x)=1

Please help, how to solve this equation - Aug 29th 2012, 01:11 PMHallsofIvyRe: Trigonometric Equation Problem
I would start by using the trig identity $\displaystyle cos(2x)= cos^2(x)- sin^2(x)= (1- sin^2(x)- sin^2(x))= 1- 2sin^2(x)$

Using that, $\displaystyle 4 sin(x)- 2 cos(2x)= 4 sin(x)- 2+ 2sin^2(x)= 1$ so that the equation is the same as $\displaystyle 2sin^2(x)+ 4sin(x)- 3= 0$. If you let y= sin(x), that is $\displaystyle 2y^2+ 4y- 3= 0$. Can you solve that quadratic equation for y and then solve for x? - Aug 29th 2012, 01:29 PMTeloRe: Trigonometric Equation Problem
Thanks for answering

I solved quadratic equation 4u^2+4u-3=0 (sinx=u)

u1=1/2

u2=-1.5 (dont satisfy)

so sinx=1/2 but i need to write two separate answers, what x1 and x2 are, but i get this

x=(-1)^n * Pi/6 + 2Pin

how to write two answers from that one, which i learned? please help - Aug 29th 2012, 01:35 PMPlatoRe: Trigonometric Equation Problem
- Aug 29th 2012, 01:38 PMTeloRe: Trigonometric Equation Problem
can you please explain how did you get those answers?

- Aug 29th 2012, 01:42 PMPlatoRe: Trigonometric Equation Problem
- Aug 29th 2012, 01:54 PMskeeterRe: Trigonometric Equation Problem
are you not familiar with the unit circle?

http://www.mathpeer.com/images/trig/unit_circle.gif