Question

find the general solution of sinx-cosx=1

Answer 1

If cosx+sinx=1 then squaring both sides gives us:

cos2x+2cosxsinx+sin2x=1

Using the identities:
cos2x+sin2x=1 and sin2x=2sinxcosx

The equation can be simplified to: 1+sin2x=1

Therefore sin2x=0

The values of θ at which sinθ=0 are θ=nπwhere n is an integer.

But here θ=2x so x=nπ2 for n=0,±1,±2,±3...

However, since we squared the equation, we need to check all of these answers work in the original equation:

cos0+sin0=1+0=1

cos(π2)+sin(π2)=0+1=1

cosπ+sinπ=−1+0=−1 This solution is not valid.

cos(3π2)+sin(3π2)=0−1=−1 This solution is not valid either.

When we get to 2π the graphs repeat, so 2π,5π2,4π,9π2... are all valid, but the other solutions aren't.

This can be written as: 
x=π2+2πnorx=2πn where n=0,±1,±2,±3...

Answer 2

Answer:

Step-by-step explanation: If cosx+sinx=1 then squaring both sides gives us:

cos2x+2cosxsinx+sin2x=1

Using the identities:

cos2x+sin2x=1 and sin2x=2sinxcosx

The equation can be simplified to: 1+sin2x=1

Therefore sin2x=0

The values of θ at which sinθ=0 are θ=nπwhere n is an integer.

But here θ=2x so x=nπ2 for n=0,±1,±2,±3...

However, since we squared the equation, we need to check all of these answers work in the original equation:

cos0+sin0=1+0=1

cos(π2)+sin(π2)=0+1=1

cosπ+sinπ=−1+0=−1 This solution is not valid.

cos(3π2)+sin(3π2)=0−1=−1 This solution is not valid either.

When we get to 2π the graphs repeat, so 2π,5π2,4π,9π2... are all valid, but the other solutions aren't.

This can be written as:

x=π2+2πnorx=2πn where n=0,±1,±2,±3...