2 \sin( \alpha ) - \cos( \alpha ) = 2 \\ find the value of: \sin( \alpha ) + 2 \cos( \alpha ) ​

Question

`2 \sin( \alpha ) - \cos( \alpha ) = 2 \\`

find the value of:

`\sin( \alpha ) + 2 \cos( \alpha )`
​

Answer 1

`2\sin\alpha-\cos\alpha=2`

Consider the substitution
`\tan\frac\alpha2=\beta`. Then by the double angle identities we get

`\sin\alpha=2\sin\frac\alpha2\cos\frac\alpha2`

`\cos\alpha=\cos^2\frac\alpha2-\sin^2\frac\alpha2`

We also have

`\tan\frac\alpha2=\beta\implies\begin{cases}\sin\frac\alpha2=\frac\beta{√(1+\beta^2)}\\\\\cos\frac\alpha2=\frac1{√(1+\beta^2)}\end{cases}`

so that

`\sin\alpha=(2\beta^2)/(1+\beta^2)`

`\cos\alpha=(1-\beta^2)/(1+\beta^2)`

and the original equation has been transformed to

`(4\beta^2-(1-\beta^2))/(1+\beta^2)=2`

Solve for
`\beta`:

`5\beta^2-1=2+2\beta^2`

`3\beta^2=3`

`\beta^2=1`

`\beta=\pm1`

Solving for
`\alpha` gives

`\tan\frac\alpha2=-1\implies\frac\alpha2=-\frac\pi4+n\pi\implies\alpha=-\frac\pi2+2n\pi`

`\tan\frac\alpha2=1\implies\frac\alpha2=\frac\pi4+n\pi\implies\alpha=\frac\pi2+2n\pi`

where
`n` is any integer. Both
`\sin` and
`\cos` are
`2\pi`-periodic, which is to say

`\cos(x+2n\pi)=\cos x`

`\sin(x+2n\pi)=\sin x`

so that

`\sin\alpha=\sin\left(\pm\frac\pi2+2n\pi\right)=\sin\left(\pm\frac\pi2\right)=\pm1`

`\cos\alpha=\cos\left(\pm\frac\pi2+2n\pi\right)=\cos\left(\pm\frac\pi2\right)=0`

and we find that

`\sin\alpha+2\cos\alpha=\pm1`