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.

Bạn đang xem: More complicated equations and inequalities

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:

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


Thanks for contributing an answer to slovenija-expo2000.comematics Stack Exchange!

Please be sure lớn answer the question. Provide details and share your research!

But avoid

Asking for help, clarification, or responding lớn other answers.Making statements based on opinion; back them up with references or personal experience.

Use slovenija-expo2000.comJax to format equations. slovenija-expo2000.comJax reference.

Xem thêm: Tài Liệu Ôn Tập Học Kì 1 Toán 12, Tổng Hợp Kiến Thức Toán Lớp 12 Chương 1 Chọn Lọc

To learn more, see our tips on writing great answers.


Post Your Answer Discard

By clicking “Post Your Answer”, you agree lớn our terms of service, privacy policy and cookie policy


Not the answer you're looking for? Browse other questions tagged trigonometry or ask your own question.


If $t = an (x/2)$, find expressions for $sin x, cos x$ in terms of $t$. Hence, solve the equation $3sin x - 4cos x = 2$.
Site kiến thiết / biểu tượng logo © 2022 Stack Exchange Inc; user contributions licensed under cc by-sa. Rev2022.4.20.41994


Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device & disclose information in accordance with our Cookie Policy.