Solve the equation for x

Question

y = ((√3)/(2√x))+2sinx = 0

y = -((√3)/(4x^3/2)+2cosx = 0

y = cos2x = 0

y = -sin(2x)•(2) = 0

Answer 1

Let's assume that the text of the last comment is correct.

Then y = cos(2x) is one problem and its derivative is:

y' = -sin(2x)(2)

y' = -2sin(2x).

To find the critical numbers set the 1st derivative equal to zero and solve for x.

-2sin(2x) = 0

sin(2x) = 0

2x = sin^-1(0)

2x = 0, π, 2π,

OR:

2x = 0 +kπ, where k is any integer

x = kπ/2, where k is any integer

Here is a partial list (where -4 ≤ k ≤4

-2π, -3π/2, -π, -π/2, 0, π/2, π, 3π/2, 2π

This graph shows that at those points the tangent lines are horizontal.

desmos.com/calculator/vg5fr4ruwx

For the 1st problem is you set the derivative equal to zero:

(-√3/4)x^-3/2 + 2cosx = 0

Multiply every term by 4 x^3/2

-√3 + 8x^3/2cos(x) = 0

x^3/2cos(x) = √3/8

This cannot be solved for x algebraically. Likely you have not yet learned Newton's method. Visit this graph and try to make some sense of it:

desmos.com/calculator/aoh1df2nkh