Use trigonometric identities to solve each equation for the
given domain.
sin 2θ = -cos θ for 0 ≤ θ < 2π
*Anyone who has any idea which identity to use is welcome to help
Expand sin2θ = 2sinθcosθ
so
sin2θ=-cosθ
2sinθcosθ + cosθ = 0
factoring out cosθ,
cosθ(2sinθ+1)=0
cosθ(sinθ+1/2)=0
Using the zero product property, we deduce that
cosθ =0 OR
sin&theta=-1/2
Solve these two fundamental equations to give the solution 0≤θ<2π.
To solve the equation sin 2θ = -cos θ, we can use the trigonometric Pythagorean identity: sin^2(θ) + cos^2(θ) = 1.
First, let's rewrite the equation sin 2θ = -cos θ in terms of sin and cos using a double angle identity: sin 2θ = 2sin θ cos θ.
Now, substitute this back into the equation: 2sin θ cos θ = -cos θ.
Rearrange the equation to isolate one of the trigonometric functions: 2sin θ cos θ + cos θ = 0.
Factor out a common factor of cos θ: cos θ(2sin θ + 1) = 0.
Now, we can set each factor equal to zero and solve for θ within the given domain:
cos θ = 0
For this equation to be true, θ must be equal to π/2 or 3π/2, as these are the values where cos θ equals zero within the given domain 0 ≤ θ < 2π.
2sin θ + 1 = 0
Rearrange and solve for sin θ: sin θ = -1/2.
For this equation to be true, there are two possible solutions within the given domain based on the unit circle. These solutions occur at θ = 7π/6 and θ = 11π/6.
Thus, the solutions to the given equation sin 2θ = -cos θ within the domain 0 ≤ θ < 2π are:
θ = π/2, 3π/2, 7π/6, and 11π/6.