solve the equation
2 tan^2(theta)sin(theta)-tan^2(theta)=0
interval[0,360)
2 tan^2(theta)sin(theta)-tan^2(theta)=0 , I will use x for theta
tan^2 x(2sinx - 1) = 0
tan^2x = 0 or sinx = 1/2
tanx = 0 or sinx = 1/2
x = 0, 180, 360 or x = 30, 150
To solve the equation 2 tan^2(theta)sin(theta) - tan^2(theta) = 0 on the interval [0, 360), we can use some algebraic manipulations. Let's go step by step.
First, we notice that the equation contains a common factor, which is tan^2(theta). We can factor it out:
tan^2(theta) * (2sin(theta) - 1) = 0
Now, we have two factors, and for the equation to be true, at least one of the factors must equal 0. So we set each factor equal to 0 and solve for theta separately.
Factor 1: tan^2(theta) = 0
To solve this equation, we take the square root of both sides:
tan(theta) = 0
Now, we need to find the values of theta where the tangent function equals 0. For the given interval [0, 360), the tangent function is equal to 0 at theta = 0 degrees and theta = 180 degrees.
Factor 2: 2sin(theta) - 1 = 0
To solve this equation, we isolate sin(theta) by adding 1 to both sides:
2sin(theta) = 1
Next, divide both sides by 2:
sin(theta) = 1/2
Now, we need to find the values of theta where the sine function equals 1/2. From the unit circle or the reference angles, we know that sin(theta) = 1/2 when theta = 30 degrees and theta = 150 degrees.
Therefore, the solutions to the equation 2 tan^2(theta)sin(theta) - tan^2(theta) = 0, on the interval [0, 360), are theta = 0 degrees, 30 degrees, 150 degrees, 180 degrees.