The number of values of 0 = [0, π] satisfying the equation
sin6theta + sin4theta + sin2theta = 0 is
Using the sum-to-product formulas,
sin6θ + sin2θ = 2 sin4θ cos2θ, so we have
sin6θ + sin4θ + sin2θ = 0
2sin4θ cos2θ + sin4θ = 0
sin4θ (2cos2θ + 1) = 0
sin4θ = 0
cos2θ = -1/2
now finish it off. Post your work if you get stuck.
I got n = 0 , 1 which satisfy the given equation
I have no idea what n=0,1 has to do with values for θ
if sin4θ = 0, then
θ = 0+n*2π/4 , π + n*2π/4 = 0, π/4, π/2, 3π/4, π
if cos2θ = -1/2, then
2θ = 2π/3 + n*2π, 4π/3 + n*2π, so
θ = π/3, 2π/3
I mean total number of values that satisfy given equation is 2. As theta belongs to interval 0,2pi
Am i right or wrong??
To find the number of values of θ in the interval [0, π] that satisfy the equation sin^6θ + sin^4θ + sin^2θ = 0, we can follow these steps:
Step 1: Simplify the equation to a quadratic equation:
Let's substitute sin^2θ = x. Then the equation becomes x^3 + x^2 + x = 0.
Step 2: Factor the quadratic equation:
By factoring out the common factor x, we have x(x^2 + x + 1) = 0.
Step 3: Solve for x:
Since x = sin^2θ, it must be between 0 and 1. The equation x(x^2 + x + 1) = 0 holds true if either x = 0 or x^2 + x + 1 = 0.
If x = 0, then sin^2θ = 0, which means θ = 0.
Now let's focus on solving x^2 + x + 1 = 0. We can use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / 2a,
where a = 1, b = 1, and c = 1.
Plugging in the values, we have:
x = [-(1) ± √((1)^2 - 4(1)(1))] / 2(1),
x = [-1 ± √(-3)] / 2.
Since the discriminant is negative, there are no real solutions for x in this case. Therefore, x^2 + x + 1 = 0 has no solutions.
Step 4: Find θ values:
From our analysis, we found that x = 0 satisfies the equation sin^6θ + sin^4θ + sin^2θ = 0. This means that sin^2θ = 0, which leads to sinθ = 0.
In the interval [0, π], the values of θ satisfying sinθ = 0 are θ = 0 and θ = π.
So, there are two values of θ in the interval [0, π] that satisfy the given equation.