How do you solve $sinx+cosx=1$ ?

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How do you solve $sinx+cosx=1$ ?

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Answer 1 · 2016-12-28T16:20:45

The answer is $S={2kpi,pi/2+2kpi}$ , $k in ZZ$

Explanation:

We need

$sin(A+B)=sinAcosB+sinBcosA$

$sin^2A+cos^2A=1$

We compare this equation to

$rsin(x+a)=1$

$rsinxcosa+rcosxsina=1$

$sinx+cosx=1$

Therefore,

$rcosa=1$ and $rsina=1$

So,

$cos^2a+sin^2a=1/r^2+1/r^2=2/r^2=1$

$r^2=2$ , $=>,$ r=sqrt2#

and $tana=1$ , $=>$ , $a=pi/4$

Therefore,

$sqrt2sin(x+pi/4)=1$

$sin(x+pi/4)=1/sqrt2$

$x+pi/4=pi/4+2kpi$ , $=>$ , $x=2kpi$

and

$x+pi/4=3pi/4+2kpi$ , $=>$ , $x=pi/2+2kpi$

The solutions are $S={2kpi,pi/2+2kpi}$ , $k in ZZ$

Answer 2 · 2016-12-28T16:33:47

Here's an alternative answer. Square both sides.

$(sinx + cosx)^2 = 1^2$

$sin^2x + 2sinxcosx + cos^2x = 1$

Use the identity $sin^2theta + cos^2theta = 1$ .

$1 + 2sinxcosx = 1$

$2sinxcosx = 0$

Use the identity $2sinthetacostheta = sin2theta$ :

$sin2x = 0$

$2x = 0, pi$

$x = 2pin, pi/2 + 2pin$ , where $n$ is an integer.

Hopefully this helps!

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