Question

Please solve it.. step by step..

Answer 1

Answer:

Step-by-step explanation:

c-√(a²+b²) <= acosx+bsinx+c <= c+√(a²+b²)

─► -√(a²+b²) <= acosx+bsinx <= √(a²+b²)

─►  │acosx+bsinx │ <= √(a²+b²)

let, a=ysinΘ, b=ycosΘ, so, a²+b²=y²(sin²Θ+cos²Θ)=y²

─►  │ysinΘcosx+ycosΘsinx │ <= √(a²+b²)

─►  │ysin(Θ+x) │ <= √(a²+b²)

─►  │ysin(Θ+x) │ <= √y²

─►  │sin(Θ+x) │ <= 1 ..............................(i)

for any value of (Θ+x)  ..(i)  is  true

so, (Θ+x)  belongs to R, where Θ = tan^(-1) (a/b)

x belongs to R -tan^(-1) (a/b)

we can write x belongs to R