Solve 2sin2θ = 1 + cos2θ for 0° ≤ θ ≤ 180°
sin2θ = 2sinθcosθ (double angle formula for sine)cos2θ = cos2θ - sin2θ (double angle formula for cosine) = 2cos2θ - 1 (utilising the trignometric identity sin2θ + cos2θ = 1)
We substitute these into our original equation to get4sinθcosθ = 1 + 2cos2θ - 1
Rearranging we getcosθ(cosθ - 2sinθ) = 0
Therefore, either cosθ = 0 or cosθ - 2sinθ = 0.
cosθ = 0 when θ = 90° only in the range we are given (draw a diagram)
We can divide the right hand equation by cosθ (note this is only valid when cosθ ≠ 0 <--> when θ ≠ 90°)This gives us1 - 2tanθ = 0 (Use inequality tanθ = sinθ/cosθ)
Which rearrages totanθ = 1/2Soθ = 26.6° only (draw diagram)(Note θ ≠ 90° so this was all valid)
We have θ = 26.6° and 90° as solutions.
SN
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