cos(x) + sin(x) = 2 | Solve for x

Added: 2026-05-14

[00:00:00]Today we want to find the value of X given the cosine of X plus the sine of X equals two.

[00:00:08]Pause this video, try to solve it on your own, and let me know your answer in the comments.

[00:00:14]All right, let's begin with an important observation. If you plot the graph of this function, we find that for real values of X, the maximum value that cosine of X plus sine of X can ever reach is square root of two. So, if we are trying to make this sum equal to two, it is simply not possible in the real world. That immediately tells us something interesting that if a solution exists, it must come from the complex world. So now, in order to solve this problem, we will step into that complex world, and this is where the magic of the goat mathematician Euler comes in.

[00:00:52]Because of Euler, we can express cosine and sine entirely in terms of Euler number E and imaginary unit I, which makes the problem much more algebraic and manageable. So, we rewrite cosine of X as E raised to IX plus E raised to -IX divided by two, and sine of X as E raised to IX minus E raised to -IX divided by 2i. After substituting these into the equation, everything is now expressed in terms of exponentials.

[00:01:28]To simplify things further, we introduce a new variable. Let Y be equal to E raised to IX. This is a clever move because now E raised to -I * X becomes 1 / Y. So, the equation becomes Y + 1 / Y divided by 2 + Y - 1 / Y divided by 2i = 2.

[00:01:54]Great. The next goal is to eliminate the fraction. So, we multiply the entire equation by 2i.

[00:02:01]Now, expand it step by step. This becomes I * Y ^ 2. Then this becomes + I. Then this part becomes + Y ^ 2, and finally, this part becomes -1.

[00:02:13]The right side becomes 4iY.

[00:02:16]Nice. Now, add both these Y ^ 2 terms to get 1 + I * Y ^ 2. Then take this on the left-hand side to get -4iY plus add both these constant parts to get I -1, and that equals zero. Oh my god, how beautifully we have converted a trigonometric equation to an exponential equation to a simple standard quadratic equation.

[00:02:44]You know what? There is a nice way to make the coefficients look much cleaner.

[00:02:49]For that, simply divide the entire expression by 1 + I. We get Y ^ 2 -4i / 1 + I + I -1 / 1 + I = 0. Now, simplify them. First, take -4i / 1 + I. To remove I from the denominator, multiply both numerator and denominator by 1 - I.

[00:03:15]The denominator becomes 1 - I ^ 2, and since I ^ 2 is -1, this becomes 1 + 1, which is two. Now, expand the numerator.

[00:03:27]-4i * 1 gives -4i, and -4i * -I gives +4i ^ 2, which becomes -4 because I ^ 2 is -1. Dividing everything by two gives -2i -2. Now, move to the second fraction, I -1 / 1 + I. Again, multiply numerator and denominator by 1 - I. The denominator becomes two, same as before. Expand the numerator. We get I * 1 gives I, I * -I gives -I ^ 2, which becomes +1. -1 * 1 gives -1, and -1 * -I gives +I.

[00:04:15]Combining terms, we get 2i. Dividing by two gives just I. Wow. So now, simply substitute both simplified results back into the equation. The expression becomes Y ^ 2 -2 * I + 1 * Y + I = 0.

[00:04:34]Now, we can use the quadratic formula to find the value of Y. Here, A is 1, B is this, and C is I. This is the formula.

[00:04:44]-B is this. Now, compute B ^ 2. That is this, which becomes 4 * I + 1 ^ 2.

[00:04:53]Expanding I + 1 ^ 2 gives I ^ 2 + 2i + 1. Since I ^ 2 is -1, this becomes -1 + 2i + 1, and the -1 and +1 cancel.

[00:05:06]Leaving 2i. So, B ^ 2 becomes 4 * 2i, which is 8i. Next, compute 4AC.

[00:05:16]Since A is 1 and C is I, this becomes 4i.

[00:05:21]Now, look at the discriminant. That is B ^ 2 -4AC, which is 8i -4i, giving 4i. So, the square root part becomes the square root of 4i, which simplifies to 2 * square root of I.

[00:05:38]Finally, divide everything by two.

[00:05:41]This gives Y = I plus one plus or minus square root of I.

[00:05:48]Okay, now comes another interesting part. What exactly is the square root of I? Using Euler's formula again, E raised to I * pi / 2 becomes cosine pi / 2 + I sine pi / 2. Cosine pi / 2 is zero, and sine pi / 2 is one. Thus, E raised to I * pi / 2 equals I.

[00:06:13]Therefore, the square root of I equals I raised to half, or E raised to I * pi / 2 whole raised to half, which equals E raised to I * pi / 4. Great. Now, E raised to I * pi divided by four can be rewritten in terms of cosine and sine like this.

[00:06:36]Both cosine and sine pi / 4 equals 1 / root 2. And thus, E raised to I * pi divided by four, or root I, becomes 1 / root 2 * 1 + I. This means 1 + I can be written as root 2 * E raised to I pi / 4. So, do one thing. In this expression for Y, write this 1 + I as root 2 * E raised to I pi / 4, and write this square root of I as E raised to I pi / 4. So, Y becomes root 2 plus or minus 1 * E raised to I pi / 4. Amazing. Do one last thing. Note that we can rewrite any number K as E raised to natural log of K. So, rewrite this thing as E raised to natural log of root 2 plus or minus 1.

[00:07:34]So, after combining both of them, we get E raised to I pi / 4 + log of root 2 plus or minus 1. We can also add +2 pi * m i because exponentials in the complex world are periodic.

[00:07:50]Now, remember that we defined Y as E raised to I * X. So now, we equate this with the expression we just found.

[00:07:59]Since the base is the same for both, which is E, thus, we can also equate powers. We get IX equals this. Next, divide everything by I.

[00:08:11]While doing this, we use the fact that 1 / I is -I, which helps simplify the expression further. So, in the end, we get a complete complex solution for X, which includes a real part involving pi divided by four, and also periodic multiples of 2 pi, and an imaginary part involving this logarithm.

[00:08:36]That was a super fun problem, wasn't it?

[00:08:38]If you enjoyed this video, please don't forget to like, share, and subscribe to our channel. So good.

[00:08:51]>> [music]