sinax +cosx=1/√2 than find genral solution
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x=π2+2πnorx=2πn where n=0,±1,±2,±3...
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...
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...
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