How do you solve sin2θ−cosθ=0 between -pi/2 less than or equal to θ less than or equal to 3pi/2?
sin2θ−cosθ=0
2sinØcosØ - cosØ = 0
cosØ(2sinØ - 1)
cosØ = 0 or sinØ = 1/2
from cosØ = 0 , where does the cosine curve cut the x-axis in your domain?
http://www.wolframalpha.com/input/?i=y+%3D+cosx
where is sinØ = 1/2 in your domain?
http://www.wolframalpha.com/input/?i=y+%3D+sinx+,+y+%3D+1%2F2
Thank you so much.
To solve the equation sin(2θ) - cos(θ) = 0, we need to find the values of θ that make the equation true.
Let's break down the equation and solve it step by step:
Step 1: Use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to manipulate the equation.
sin(2θ) - cos(θ) = 0
2sin(θ)cos(θ) - cos(θ) = 0
cos(θ)(2sin(θ) - 1) = 0
Step 2: Set each factor equal to zero and solve for θ.
cos(θ) = 0
θ = π/2 + kπ, where k is an integer
2sin(θ) - 1 = 0
sin(θ) = 1/2
θ = π/6 + 2kπ or θ = 5π/6 + 2kπ, where k is an integer
Step 3: Combine all the solutions obtained from both factors.
θ = π/2 + kπ, where k is an integer
θ = π/6 + 2kπ, where k is an integer
θ = 5π/6 + 2kπ, where k is an integer
So, the solutions for the given equation in the specified range -π/2 ≤ θ ≤ 3π/2 are:
θ = π/2, π/6, 5π/6