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Question-18723




Question Number 18723 by mondodotto@gmail.com last updated on 28/Jul/17
Answered by Tinkutara last updated on 29/Jul/17
2sin2θ + 15cos2θ = 10  ((2sin2θ)/( (√(229)))) + ((15cos2θ)/( (√(229)))) = ((10)/( (√(229))))  cos(2θ − α) = ((10)/( (√(229)))) , where α = cos^(−1) (((15)/( (√(229)))))  θ = nπ ± (5/( (√(229)))) + (1/2)cos^(−1) (((15)/( (√(229)))))
$$\mathrm{2sin2}\theta\:+\:\mathrm{15cos2}\theta\:=\:\mathrm{10} \\ $$$$\frac{\mathrm{2sin2}\theta}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{15cos2}\theta}{\:\sqrt{\mathrm{229}}}\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}} \\ $$$$\mathrm{cos}\left(\mathrm{2}\theta\:−\:\alpha\right)\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}}\:,\:\mathrm{where}\:\alpha\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right) \\ $$$$\theta\:=\:{n}\pi\:\pm\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right) \\ $$

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