The Number Of Solutions Of

     
I"m trying 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 ?


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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.


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Another way khổng lồ solve this is to lớn write $cos x = (e^ix+e^-ix)/2$ & $sin x = (e^ix-e^-ix)/2i$. 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 $n$.

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Following where you got stuck và squaring both sides & 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 $$2x = frac 3pi2+2kpi Rightarrow x=frac 3pi4+kpi$$


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I"m going to lớn go through this assuming that you"re solving for solutions within $<0,2pi>$.

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

With this, we can pull out our trusty old unit circle:

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Then, we need to lớn find any angles on the circle where $cos x = -sin x$

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Sorry for the low res on the second image. But, as you can see, we have our angles. The solutions to lớn $sin x+cos x=0$ between $<0,2pi>$ are $frac3pi4$ và $frac7pi4$.

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$$x+(pi/4)=npi (ninconhantaohpg.combb Z).$$


Hint:

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

$cos x=cosalpha$ gives $x=pmalpha+2kpi$ for $kinconhantaohpg.combbZ$


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 $$x=frac3π4+πk,kinconhantaohpg.comrm Z.$$


$$cos x + sin x = 0$$

Multiply $dfracsqrt22$ to 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:

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


If $t = an (x/2)$, find expressions for $sin x, cos x$ in terms of $t$. Hence, solve the equation $3sin x - 4cos x = 2$.
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