Trigonometric Equation Problem

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August 29th 2012, 12:48 PM
Telo

Trigonometric Equation Problem

4sinx-2cos(2x)=1

Please help, how to solve this equation
August 29th 2012, 01:11 PM
HallsofIvy

Re: Trigonometric Equation Problem

I would start by using the trig identity $cos(2x)= cos^2(x)- sin^2(x)= (1- sin^2(x)- sin^2(x))= 1- 2sin^2(x)$

Using that, $4 sin(x)- 2 cos(2x)= 4 sin(x)- 2+ 2sin^2(x)= 1$ so that the equation is the same as $2sin^2(x)+ 4sin(x)- 3= 0$ . If you let y= sin(x), that is $2y^2+ 4y- 3= 0$ . Can you solve that quadratic equation for y and then solve for x?
August 29th 2012, 01:29 PM
Telo

Re: 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
August 29th 2012, 01:35 PM
Plato

Re: Trigonometric Equation Problem

Quote:

Originally Posted by Telo

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

Actually the two principal values are $\frac{\pi}{6}~\&~\frac{5\pi}{6}$
August 29th 2012, 01:38 PM
Telo

Re: Trigonometric Equation Problem

can you please explain how did you get those answers?
August 29th 2012, 01:42 PM
Plato

Re: Trigonometric Equation Problem

Quote:

Originally Posted by Telo

can you please explain how did you get those answers?

$\sin\left(\frac{\pi}{6}\right)=0.5~\&~\sin\left( \frac{5\pi}{6}\right)=0.5$
August 29th 2012, 01:54 PM
skeeter

Re: Trigonometric Equation Problem

Quote:

Originally Posted by Telo

can you please explain how did you get those answers?

are you not familiar with the unit circle?

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

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