Question

Find the general solution of sinX - cosX = 1

Answer 1

Step-by-step explanation:

sinX-cosX=1

⇒(sinX-cosX)²=1²

⇒sin²X+cos²X-2sinXcosX=1

⇒1-sin²X=1  

[(sin²X+cos²X)=1, and sin²X=(2sinXcosX)]

⇒sin²X=0

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

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

Now, we need to check....

sin0-cos0=1-0=1

sin(π²)-cos(π²)=1-0= 1

sinπ-cosπ=−1-0=−1

This solution is not valid.

sin(3π2)-cos(3π2)=-1+0=−1

This solution is also not valid.

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...