What values for theta(0<_theta<_2pi) satisfy the equation?
3sin theta = sin theta-1
3sinθ = sinθ - 1
I suspect a typo, but as things stand,
2sinθ = -1
sinθ = -1/2
since sin π/6 = 1/2, and sinθ < 0 in QII,QIV, we have
θ = π+π/6 and 2π-π/6
If you are looking at solving
3sin(θ)=sin-1(θ)
then you would be solving
f(x)=3sin(x)-asin(x)=0
in which case the trivial solution x=0 can be obtained by inspection.
If you're looking for
3sin(θ)=1/sin(θ)
then there are 4 solutions between 0 and 2&pi.
Transform the equation into a quadratic equation in s=sin(θ)
then the equation becomes
f(s)=3s²-1=0 where s≠0
so s=±sqrt(1/3), meaning 4 roots.
I guess Steve's interpretation of the question is correct. There was no indication of power!
Please go with Steve's solution.
To find the values of theta that satisfy the equation 3sin(theta) = sin(theta) - 1, we can use some algebraic manipulations. Here's how you can do it:
Step 1: Start with the given equation: 3sin(theta) = sin(theta) - 1.
Step 2: Simplify the equation by combining like terms: 3sin(theta) - sin(theta) = -1.
Step 3: Combine the terms on the left-hand side: 2sin(theta) = -1.
Step 4: Divide both sides of the equation by 2 to isolate the sine term: sin(theta) = -1/2.
Step 5: Now, we need to find the values of theta that satisfy the equation sin(theta) = -1/2.
Step 6: Recall the unit circle and the special values of sine. We know that sine is negative (i.e., below the x-axis) in two quadrants: the third quadrant (180° to 270°) and the fourth quadrant (270° to 360°).
Step 7: Determine the angles in these quadrants where sin(theta) is equal to -1/2. The special angle in the third quadrant is 7π/6, and in the fourth quadrant, it is 11π/6.
Step 8: Since theta can be any angle between 0° and 360° (or 0 and 2π), the solutions are theta = 7π/6 and theta = 11π/6.
Therefore, the values for theta between 0 and 2π that satisfy the equation 3sin(theta) = sin(theta) - 1 are theta = 7π/6 and theta = 11π/6.