find the general solution for the equation:
sin2x+ square root 3cosx =0
write it this way:
sin (2x) + √3 cosx = 0
2sinxcosx + √3cosx = 0
cosx(2sinx + √3) = 0
sinx = 0 or sinx = - √3/2
if sinx = 0
x = 0, π, 2π, ... ------> kπ , where k is an integer
if sinx = -√3/2
I know that sin30° = sin π/6 = +√3/2 , but x is in quadrants III or IV
so x = 180+30 = 210° = 7π/6
or x = 360-30 = 330° = 11π/6
the period of sinx is 2π
general :
x = 7π/6 + 2πk
x = 11π/6 + 2πl
x = 2πk
To find the general solution for the equation sin(2x) + √3cos(x) = 0, we can use the identity sin(2x) = 2sin(x)cos(x).
Let's rewrite the equation using this identity:
2sin(x)cos(x) + √3cos(x) = 0
Now, we can factor out cos(x):
cos(x)(2sin(x) + √3) = 0
To find the values of x that satisfy this equation, we need to consider the two possible cases:
Case 1: cos(x) = 0
In this case, x can be any angle that gives us a cosine value of 0. This happens at x = π/2 + nπ and x = 3π/2 + nπ, where n is an integer.
Case 2: 2sin(x) + √3 = 0
Let's solve this equation separately:
2sin(x) = -√3
sin(x) = -√3/2
This equation is satisfied when x = 7π/6 + 2nπ and x = 11π/6 + 2nπ, where n is an integer.
So, the general solution for the equation sin(2x) + √3cos(x) = 0 is:
x = π/2 + nπ, x = 3π/2 + nπ, x = 7π/6 + 2nπ, and x = 11π/6 + 2nπ, where n is an integer.