To solve the equation sqrt(3)sin^2x + sinx*cosx = 0, we can first factor out sinx from the equation:
sinx(sqrt(3)sinx + cosx) = 0
Now we have two possible solutions:
For sinx = 0, the solution is x = nπ, where n is an integer.
For the second equation sqrt(3)sinx + cosx = 0, we can square both sides to get rid of the square root:
(√3sinx)^2 + 2√3sinxcosx + (cosx)^2 = 03sin^2x + 2√3sinxcosx + cos^2x = 0
Now we have a quadratic equation in terms of sinx and cosx. Let's substitute sinx = a and cosx = b:
3a^2 + 2√3ab + b^2 = 0(3a + b)(a + √3b) = 0
This equation has two possible solutions:
Plugging back sinx = a and cosx = b, we get:
3sinx + cosx = 03sinx = -cosxtanx = -1/3x = arctan(-1/3)
sinx + √3cosx = 0sinx = -√3cosxtanx = -√3x = arctan(-√3)
Therefore, the solutions to the equation sqrt(3)sin^2x + sinx*cosx = 0 are x = nπ, x = arctan(-1/3), and x = arctan(-√3) where n is an integer.
To solve the equation sqrt(3)sin^2x + sinx*cosx = 0, we can first factor out sinx from the equation:
sinx(sqrt(3)sinx + cosx) = 0
Now we have two possible solutions:
sinx = 0sqrt(3)sinx + cosx = 0For sinx = 0, the solution is x = nπ, where n is an integer.
For the second equation sqrt(3)sinx + cosx = 0, we can square both sides to get rid of the square root:
(√3sinx)^2 + 2√3sinxcosx + (cosx)^2 = 0
3sin^2x + 2√3sinxcosx + cos^2x = 0
Now we have a quadratic equation in terms of sinx and cosx. Let's substitute sinx = a and cosx = b:
3a^2 + 2√3ab + b^2 = 0
(3a + b)(a + √3b) = 0
This equation has two possible solutions:
3a + b = 0a + √3b = 0Plugging back sinx = a and cosx = b, we get:
3sinx + cosx = 0
3sinx = -cosx
tanx = -1/3
x = arctan(-1/3)
sinx + √3cosx = 0
sinx = -√3cosx
tanx = -√3
x = arctan(-√3)
Therefore, the solutions to the equation sqrt(3)sin^2x + sinx*cosx = 0 are x = nπ, x = arctan(-1/3), and x = arctan(-√3) where n is an integer.