I"m trying to lớn solve an equation here but unfortunately I can"t.The equation:\$\$cos x + sin x = 0\$\$I"m trying to solve this by replacing \$cos x\$ with \$(1-t^2)/(1+t^2)\$ & \$sin x\$ with \$2t/(1+t^2), t= an x/2, \$ but I can"t get the right solution.Also I have tried by squaring both sides but still nothing.

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Can anyone help me ?  Note that \$\$cos x + sin x = 0 iff cos x = -sin x\$\$

Now, \$cos x\$ cannot equal zero, since if it did, \$sin x = -1\$ or \$sin x = 1\$, in which case the given equation isn"t satisfied.

So we can divide by \$cos x\$ to get \$\$1 = dfrac-sin xcos x = - an x iff an x = -1\$\$

Solving for \$x\$ gives us the values \$x = dfrac 3pi4 + kpi\$, where \$k\$ is any integer. Another way lớn solve this is to write \$cos x = (e^ix+e^-ix)/2\$ and \$sin x = (e^ix-e^-ix)/2i\$. The equation simplifies in a couple of easy steps to lớn \$e^2ix= e^-pi i/2\$. This is equivalent to lớn \$2x= -pi /2 + 2pi n\$, so \$x= -pi /4 + pi n\$ for integral \$n\$. Following where you got stuck và squaring both sides and you obtain

\$\$sin^2 x+cos^2x+2sin x cos x=0 Rightarrow 1+sin2x=0\$\$

Using \$sin 2x = 2sin x cos x\$.

This means that

\$\$sin 2x = -1\$\$ and hence \$\$2x = frac 3pi2+2kpi Rightarrow x=frac 3pi4+kpi\$\$ I"m going to go through this assuming that you"re solving for solutions within \$<0,2pi>\$.

\$cos x+sin x=0\$\$implies cos x=-sin x\$

With this, we can pull out our trusty old unit circle: Then, we need to lớn find any angles on the circle where \$cos x = -sin x\$ Sorry for the low res on the second image. But, as you can see, we have our angles. The solutions to \$sin x+cos x=0\$ between \$<0,2pi>\$ are \$frac3pi4\$ and \$frac7pi4\$.

Hope that helps!

Since \$\$cos x+sin x=sqrt 2sin(x+(pi/4)),\$\$you can solve \$\$sin(x+(pi/4))=0.\$\$

Hence, you"ll have\$\$x+(pi/4)=npi (ninslovenija-expo2000.combb Z).\$\$

Hint:

\$cos x=-sin x=cosleft(frac12pi+x ight)\$

\$cos x=cosalpha\$ gives \$x=pmalpha+2kpi\$ for \$kinslovenija-expo2000.combbZ\$

You already have some good answers, but just for the fun of it here"s another way:

\$\$cos x+sin x=cos x+cos(π/2-x)=2cos(π/4)cos(x-π/4)=0,\$\$ which implies \$\$cos(x-π/4)=0,\$\$ so that we have \$\$x-fracπ4=fracπ2+πk,\$\$ where \$k\$ is any integer. Finally this gives \$\$x=frac3π4+πk,kinslovenija-expo2000.comrm Z.\$\$

\$\$cos x + sin x = 0\$\$

Multiply \$dfracsqrt22\$ lớn both sides:

\$\$dfracsqrt22cos x + dfracsqrt22sin x = 0\$\$

Or:

\$\$cosdfracpi4cos x + sindfracpi4sin x = 0\$\$

Then:

\$\$cosleft(x - dfracpi4 ight) = 0\$\$

So:

\$\$x - dfracpi4 = dfracpi2 + kpi\$\$

Therefore:

\$\$x = dfrac3pi4 + kpiqquad (k in Bbb Z)\$\$

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