Question

Cos(3пи/2 - 2x)=корень из 3 Sinx

Answer 1

По формуле приведения:
Cos(3пи/2 - 2x)=-sin(2x)

синус двойного угла:
sin2a=2sina*cosa


$cos(3\pi/2 - 2x)= \sqrt{3} sinx \\ -sin(2x)=\sqrt{3} sinx \\ \sqrt{3} sinx +sin(2x)=0 \\ \sqrt{3} sinx +2sinx*cosx=0 \\ sinx( \sqrt{3} +2cosx)=0 \\ \\ 1) \ sinx=0 \\ \\ x=\pi n, n \in Z \\ \\ 2) \ \sqrt{3} +2cosx=0 \\ \\ cosx=- \frac{ \sqrt{3} }{2} \\ \\ x=^+_- \frac{5 \pi }{6}+2 \pi n, n \in Z \\ \\ \\ OTBET: \ \pi n; \ \frac{5 \pi }{6}+2 \pi n; \ - \frac{5 \pi }{6}+2 \pi n, n \in Z$

Answer 2

Let's start by simplifying the left-hand side of the equation using the angle subtraction formula for cosine:

cos(3π/2 - 2x) = cos(3π/2)cos(2x) + sin(3π/2)sin(2x) cos(3π/2) = 0 and sin(3π/2) = -1, so we can simplify further:

cos(3π/2 - 2x) = 0*cos(2x) + (-1)*sin(2x) = -sin(2x)

Now we can rewrite the equation as:

-sin(2x) = √3*sin(x)

Dividing both sides by sin(x) (which we assume is not zero), we get:

-tan(2x) = √3

Taking the arctangent of both sides, we get:

2x = atan(-√3) + kπ, where k is an integer x = (atan(-√3) + kπ)/2, where k is an integer

Note that the arctangent of -√3 is -π/3, so we can simplify further:

x = (-π/6 + kπ)/2, where k is an integer

Therefore, the general solution for the equation is:

x = -π/12, π/4, 7π/12, 3π/4, 19π/12, 5π/4, ...

where we have substituted k = 0, 1, -1, 2, -2, 3, -3, ... into the formula.