Question Number 115062 by Sudip last updated on 23/Sep/20

If  2cosθ−sinθ=(1/( (√2)))  (0°<θ<90°)  then  2sinθ+cosθ= ¿

Answered by PRITHWISH SEN 2 last updated on 23/Sep/20

let 2sin θ+cos θ = k ....(ii)  then by solving eqn (i)(given) & (ii) we get  sin θ = (1/5)(2k−(1/( (√2)))) & cos θ= (1/5)(k+(√2))  from sin ^2 θ+cos ^2 θ=1 we get  k=±(3/( (√2)))  as 0<θ<90  ∴ 2sin 𝛉+cos θ = (3/( (√2)))

Answered by Dwaipayan Shikari last updated on 23/Sep/20

4cos^2 θ+sin^2 θ−4sinθcosθ=((sin^2 θ+cos^2 θ)/2)  8cos^2 θ+2sin^2 θ−8sinθcosθ=sin^2 θ+cos^2 θ  7cos^2 θ+sin^2 θ−8sinθcosθ=0  7cos^2 θ−7sinθcosθ−cosθsinθ+sin^2 θ=0  7cosθ(cosθ−sinθ)−sinθ(cosθ−sinθ)=0  cosθ=sinθ   ((π/4),−(π/4))  2sinθ+cosθ=±(3/( (√2)))  or  θ=tan^(−1) (1/7)