Question

Sinx+2cosx=1 then prove that 2sinx-cosx=2

Answer 1

Given:

$sin\,x+2\,cos\,x=1$

Squaring on both sides, we get

$(sin\,x+2\,cos\,x)^2=1$

Using the algebraic identity

$\boxed{\bf(a+b)^2=a^2+b^2+2ab}$

$sin^2x+4cos^2x+4sin\,x\,cos\,x=1$

Using

$\boxed{\bf\,sin^2\theta+cos^2\theta=1}$

$sin^2x+4(1-sin^2x)+4sin\,x\,cos\,x=sin^2x+cos^2x$

$4-4sin^2x+4sin\,x\,cos\,x=cos^2x$

$4=cos^2x+4sin^2x-4sin\,x\,cos\,x$

$4=cos^2x+(2sinx)^2-2\,cos\,x(2\,sin\,x)$

$4=(cosx+2sinx)^2$

Taking square root on both sides, we get

$2=cosx+2sinx$

$\implies\boxed{\bf\,cosx+2sinx=2}$

Answer 2

Answer:

You are most welcome!!

Step-by-step explanation: