2sin2x+1=0
sin2x = -1/2
2x = 210° or 330°
x = 105° or 165°
the period of sin2x is 180°, so two other answers are
105+180 = 285° or 165+180 = 345°
in degrees: x = 105° 165° 285° 345°
in radians: x = 7π/12, 11π/12,19π/12, 23π/12
To solve the equation 2sin(2x) + 1 = 0, we need to isolate the variable x.
Let's start by isolating the term containing the sine function:
2sin(2x) = -1
Next, we divide both sides of the equation by 2 to get sin(2x) by itself:
sin(2x) = -1/2
Now, we need to find the values of 2x that satisfy this equation within a given range. Usually, when solving for x in trigonometric equations, we look for solutions within a 360-degree interval or a 2π radian interval.
In this case, the sine function has a value of -1/2 at two specific angles: -30 degrees (-π/6 radians) and -150 degrees (-5π/6 radians). To find these angles, we can use the inverse sine function (also known as arcsin or sin^(-1)).
Using a calculator, we find that the inverse sine of -1/2 is -π/6 + nπ (where n is an integer) and -5π/6 + nπ. These angles will give us the values of 2x that satisfy the equation.
So, we have two possible solutions:
1. 2x = -π/6 + nπ
2. 2x = -5π/6 + nπ
To find the values of x, we divide both sides of each equation by 2:
1. x = (-π/6 + nπ)/2
2. x = (-5π/6 + nπ)/2
Depending on the range or interval specified in the problem, you can substitute different values for n to get different values of x that satisfy the equation.