solving triq equation?

Question

2sin2x-2cosx-sinx=1

I don't see a way to factor or cancel out anything. any help will be appreciated!:)

if i use the identities, i only get it to:
2 cos(sinx-1)=sinx+1
And, you can't do anything with that...i think

Answer 1

where did you get THIS problem?
The (real-valued) solutions are:

3π/2 + 2πn and ≈ 2.945023353 + 2πn where n = 1,2,3,...
(radians, of course)

2sin2x - 2cosx - sinx = 1
4sin(x)cos(x) - 2cos(x) - sin(x) = 1
2cos(x)*[2sin(x) - 1] - sin(x) = 1
let y = sin(x) then cos(x) = √(1-sin^2(x)) = √(1-y^2)
so,
2√(1-y^2) *[2y - 1] - y = 1
or
2√(1-y^2) *[2y - 1] = 1 + y
squaring both sides,
4(1 - y^2)(2y - 1)^2 = (1 + y)^2
4(1 - y)(1 + y)(4y^2 - 4y + 1) = (1 + y)^2
we can divide both side by (1+y) noting that this shows that
y = -1 is a solution, or sin(x) = -1, x = 3π/2 + 2πn ; n = 1,2,3,...

then we have
4(1 - y)(4y^2 - 4y + 1) = (1 + y)
expanding,
16y^3 - 32y^2 + 21y - 3 = 0

This is why I asked where you got this problem. This cubic equation is not easily solved. There are two complex conjugate solutions as well as y ≈ 0.195305854...(I wont show a derivation. Use Mathematica or something similar. Plug it in-it works. You could also use Cardano's formula to find an exact value but I'm sure not doing that here and now. Maybe if you paid me.)

So, sin(x) ≈ 0.195305854
Then x ≈ 0.1965696 or π - 0.1965696 ≈ 2.945023353
But the first one gets rejected since cos(x) must be negative.(This spurious solution comes from squaring the original equation) Also, we can add even multiples of 2π to the solution to get another. So,
x ≈ 2.945023353 + 2πn where n = 1,2,3,...

These are the real solutions. As I indicated, there are two complex conjugate solutions which will yield additional complex solutions for x.
sin(x) ≈ 0.90234707 ± 0.3818408i

You can figure out x in these cases!!!

Answer 2

Remeber sin(2x)=2sinxcosx=
2tan(x)/1+tanx^2


2(2sinxcosx) -2cosx = sinx

4sincosx -2cosx= 1- sinx

2cosx(2sinxcosx -cosx)= 1-sinx

2cosx = 1-sinx/ (2tan(x)/ 1+ tanx^2) -cosx

2cosx= 1-sinx/ sinxcosx - cosx

2cosx= -1(sinx -1)/ cosx(sinx -1)

2cos x= -1/cos x

2cosx= -cscx

cos(x)= -csc(x)/2= -1/2cosx


the identity is false but i don't know what else you want to do after simplifying it


Who keeps giving people who answer questions thumbs doen, don't be overly zealous or jealous because you're not smart enough to do it

Answer 3

My trig is very rust but I think you can combine 2 sin2x and -sinx that should make the equation 2sinx - 2cosx = 1

Answer 4

Maybe these identities will help...
sin (2x) = 2 sin(x) cos(x)
cos (2x) = 1 - 2sin^2 (x)
sin^2 (x) + cos^2 (x) = 1