Thread: Trigonometric Equation Problem

4sinx-2cos(2x)=1

Please help, how to solve this equation

I would start by using the trig identity

Using that, so that the equation is the same as . If you let y= sin(x), that is . Can you solve that quadratic equation for y and then solve for x?

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

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

can you please explain how did you get those answers?

Originally Posted by Telo

can you please explain how did you get those answers?

Originally Posted by Telo

can you please explain how did you get those answers?

are you not familiar with the unit circle?