1. ## Trigonometric Equation Problem

4sinx-2cos(2x)=1

2. ## Re: 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?

3. ## Re: Trigonometric Equation Problem

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

4. ## Re: Trigonometric Equation Problem

Originally Posted by Telo
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 $\displaystyle \frac{\pi}{6}~\&~\frac{5\pi}{6}$

6. ## Re: Trigonometric Equation Problem

Originally Posted by Telo
$\displaystyle \sin\left(\frac{\pi}{6}\right)=0.5~\&~\sin\left( \frac{5\pi}{6}\right)=0.5$