Equations containing circular functions
and unit circle... solve
2cosx sinx + sinx = 0
To solve the equation 2cos(x)sin(x) + sin(x) = 0, we can factor out sin(x) from both terms:
sin(x)(2cos(x) + 1) = 0
Now we have two factors in the equation: sin(x) = 0 and 2cos(x) + 1 = 0.
Let's solve each of these factors separately:
1. sin(x) = 0:
To find the solutions for this factor, we need to determine the values of x that make sin(x) equal to zero. Since sin(x) is zero at x = 0, π, 2π, and so on, the general solutions for this factor are x = nπ, where n is an integer.
2. 2cos(x) + 1 = 0:
To solve this factor, we need to isolate cos(x). Subtracting 1 from both sides, we have:
2cos(x) = -1
Dividing both sides by 2, we get:
cos(x) = -1/2
To find the values of x that satisfy this equation, we can use the unit circle or the trigonometric values of special angles. From the unit circle, we know that cos(x) is -1/2 at two angles: x = 2π/3 and x = 4π/3.
So, the general solutions for this factor are x = 2π/3 + 2nπ and x = 4π/3 + 2nπ, where n is an integer.
Combining the solutions for both factors, we have the general solution to the equation 2cos(x)sin(x) + sin(x) = 0 as:
x = nπ, 2π/3 + 2nπ, and 4π/3 + 2nπ, where n is an integer.