8 Tan^2 theta sin theta + 4 Tan^2 theta = 0
fcator out 4tan^2θ to get
4tan^2θ(2sinθ+1) = 0
so, tanθ = 0 or sinθ = -1/2
θ=0 or θ = -π/6
You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be
θ = πn
θ = π/6 + +2πn
for any integer n
fcator out 4tan^2θ to get
4tan^2θ(2sinθ+1) = 0
so, tanθ = 0 or sinθ = -1/2
θ=0 or θ = 7π/6 or θ = 11π/6
You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be
θ = πn
θ = 7π/6 + +2πn
θ = 11π/6 + +2πn
for any integer n
To solve the equation 8tan^2(theta)sin(theta) + 4tan^2(theta) = 0, we will use some trigonometric identities and algebraic manipulations.
Let's factor out common terms:
4tan^2(theta) * (2sin(theta) + 1) = 0
Now we have two factors: 4tan^2(theta) = 0 and (2sin(theta) + 1) = 0.
Let's solve the first factor:
4tan^2(theta) = 0
To find the values of theta that satisfy this equation, we can set tan^2(theta) = 0.
Using the property that tan(theta) = sin(theta)/cos(theta), we have:
(sin(theta)/cos(theta))^2 = 0
Since a fraction squared can only be zero if the numerator is zero, we set sin(theta) = 0.
Thus, one solution is theta = 0.
Now let's solve the second factor:
2sin(theta) + 1 = 0
Subtracting 1 from both sides:
2sin(theta) = -1
Dividing by 2:
sin(theta) = -1/2
Looking at the unit circle or trigonometric values, we know that sin(theta) = -1/2 in the third and fourth quadrants. Therefore, the solutions for sin(theta) = -1/2 are theta = 210 degrees and theta = 330 degrees.
Combining all the solutions:
theta = 0, 210 degrees, 330 degrees.
These are the values of theta that satisfy the equation 8tan^2(theta)sin(theta) + 4tan^2(theta) = 0.