Answer: i really dont know
Step-by-step explanation: but 46
4sin x=2sin x + √3
4sin x-2sinx=√3
2sin x=√3
sinx=√3/2
x=arcsin √3/2=π/3 + 2Kπ U 2π/3+2Kπ
Sol: π/3 + 2Kπ U 2π/3+2Kπ ; K∈Z
π/3+2Kπ=60º+360ºk
2π/3+2Kπ=120º+360ºK
Answer:
Step-by-step explanation:
You want to factor out perfect squares if they exist.

Answer:
![\displaystyle Range: \\ Set-Builder\:Notation → y|1 ≥ y ≥ -1 \\ Interval\:Notation → [-1, 1] \\ \\ Domain: \\ Set-Builder\:Notation → x|x ∈ R \\ Interval\:Notation → (-∞, ∞)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Range%3A%20%5C%5C%20Set-Builder%5C%3ANotation%20%E2%86%92%20y%7C1%20%E2%89%A5%20y%20%E2%89%A5%20-1%20%5C%5C%20Interval%5C%3ANotation%20%E2%86%92%20%5B-1%2C%201%5D%20%5C%5C%20%5C%5C%20Domain%3A%20%5C%5C%20Set-Builder%5C%3ANotation%20%E2%86%92%20x%7Cx%20%E2%88%88%20R%20%5C%5C%20Interval%5C%3ANotation%20%E2%86%92%20%28-%E2%88%9E%2C%20%E2%88%9E%29)
Step-by-step explanation:
This is the graph of
in which its AMPLITUDE [<em>A</em>] ALWAYS starts ONE BLOCK ABOVE the <em>midline</em><em>.</em><em> </em>In the trigonometric formula below, −C gives the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:
![\displaystyle y = Acos[Bx - C] + D](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20Acos%5BBx%20-%20C%5D%20%2B%20D)
NOTE: Depending on how your trigonometric graphs are structured, your vertical shift [<em>D</em>] might tell you to space out the amplitude of the graphs alot more evenly on both ends.
<u>Extended</u><u> </u><u>Information on</u><u> </u><u>Trigonometric</u><u> </u><u>Graphs</u>
![\displaystyle Vertical\:Shift = D \\ Phase\:Shift\:[Horizontal\:Shift] = \frac{C}{B} \\ Period = \frac{2}{B}π \\ Amplitude = |A|](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Vertical%5C%3AShift%20%3D%20D%20%5C%5C%20Phase%5C%3AShift%5C%3A%5BHorizontal%5C%3AShift%5D%20%3D%20%5Cfrac%7BC%7D%7BB%7D%20%5C%5C%20Period%20%3D%20%5Cfrac%7B2%7D%7BB%7D%CF%80%20%5C%5C%20Amplitude%20%3D%20%7CA%7C)
I am joyous to assist you anytime.