to solve sin⁡2θ−cos⁡θ=0

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to solve sin⁡2θ−cos⁡θ=0

pattererX · Accepted Answer

To solve sin⁡2θ−cos⁡θ=0, we need to rewrite sin⁡2θ so the terms will all have angles of θ.     The double angle formula for sine is sin⁡2θ=2sin⁡θcos⁡θ so sin⁡2θ−cos⁡θ=0 can be rewritten as 2sin⁡θcos⁡θ−cos⁡θ=0.     Both terms then have a common factor of cosθcosθ that can be factored out. Factoring out cosθcosθ then gives cos⁡θ(2sin⁡θ−1)=0     The Zero Product Property states that if ab=0 ab=0, then a=0 or b=0. Therefore, if cos⁡θ(2sin⁡θ−1)=0 , then cos⁡θ=0 or 2sin⁡θ−1=0     From the unit circle, cos⁡θ=0 when θ=π2 and 3π2     Solving 2sin⁡θ−1=0 for sin⁡θ sin⁡θ gives sin⁡θ=12 . From the unit circle, sin⁡θ=12 when θ=π6 and 5π6.     The solutions of the equation on the interval [0,2π) are then x=π6, π2, 5π6, 3π2.     If you need to give ALL solutions, then you need to use the period of sine and cosine.     Both sine and cosine have a period of 2π so any value that is a multiple of 2π from the above solutions will also be solutions. That is x=π6+2πk, π2+2πk, 5π6+2πk, 3π2+2πk where k is an integer are all solutions of the equation.     Since π2 and 3π2 are π away from each other, then x=π2+2π​and x=3π2+2πk can be simplified to just x=π2+πk . All solutions to the equation are then x=π6+2πk, π2+πk, 5π6+2πk where k is an integer.

sin2θ−cosθ=0 to solve

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