General Solutions

Transcript

0:00Introduction

Hello everyone! Let us start with a simple equation: sine theta equals zero. Theta equals zero works, of course. But sine pi is also zero, and so are sine two pi and sine minus pi. So this equation has infinitely many answers. How do we write all of them in one line? That is today's topic, general solutions, and it comes up often in JEE questions. So, let's begin.

0:28Principal and general solutions

So, a solution of a trigonometric equation is a value of the variable that satisfies it. Sine, cos and tan are periodic, so one solution usually means infinitely many. With me? The solutions from zero up to two pi are called principal solutions; pi, the Greek letter, is half a turn. The square bracket means zero is included, and the round bracket means two pi is left out. Right? Let me ask you this: what are the principal solutions of tan x equals minus root three? Root three is the square root of three, the tick-shaped sign. Okay, let's check. Tan is negative in the second and fourth quadrants, with reference angle pi over three. So pi minus pi over three gives two pi over three, and two pi minus pi over three gives five pi over three. Clear? But we want every solution. The general solution is one formula for all of them, using an integer n. The curved symbol means belongs to, and the double-lined capital Z is the set of all integers. Got it?

1:32Where sine, cos and tan are zero, one or minus one

Now back to our puzzle. Theta, the Greek letter on screen, is our angle. Sine theta is zero at zero, pi, two pi, minus pi. Can you write all of them in one formula? Yes, exactly, theta equals n pi, every integer multiple of pi! And tan theta is sine theta over cos theta, so it is zero at the same places. So our puzzle is solved, okay? Now cos theta equals zero. Cos is zero at pi over two, so the answer is n pi over two, right? No, no, careful, that is a trap! At n equals two, n pi over two is pi, where cos is minus one. Cos is zero only at the odd multiples of pi over two, so theta equals two n plus one, times pi over two. Clear? Sine theta equals one happens once in each full turn, so theta equals two n pi plus pi over two. And sine theta equals minus one gives two n pi minus pi over two. With me? Cos theta equals one gives two n pi, the full turns, and cos theta equals minus one gives two n plus one, times pi, the odd multiples of pi. Right?

2:42The three main formulas

Here are the big three. Alpha, this Greek letter, is any one solution. For sine, theta equals n pi, plus minus one to the power n, times alpha. Even n gives alpha plus full turns; odd n gives odd multiples of pi, minus alpha. Sine is the same at alpha and at pi minus alpha, so one formula catches both. Okay? Now let's do cos. If alpha solves cos theta equals cos alpha, which other angle surely works too? Yes, minus alpha, because cos is an even function. And cos repeats after two pi, so theta equals two n pi plus or minus alpha; a plus on top of a minus is read plus or minus. Got it? Tan is the simplest: it repeats after just pi, so theta equals n pi plus alpha. Right? Any one solution can be alpha, but usually we take these. For sine, from minus pi over two to pi over two, ends included. For tan, the same, but with round brackets, because tan is not defined at the ends. And for cos, from zero to pi. Clear?

3:51Squares, and three traps

Now the squares; the small two means the whole sine is squared. Sine squared theta equals sine squared alpha means sine theta is plus or minus sine alpha, so alpha and minus alpha both count. Cos squared and tan squared give the same: theta equals n pi plus or minus alpha. With me? Next come the traps. If you cancel sine theta from both sides of an equation, what do you lose? Yes, exactly, you lose every solution where sine theta is zero, the whole family n pi! So never cancel a factor that can be zero. Okay? The second trap is squaring: it can add extra roots, so always check your roots in the original equation. Right? And the third trap: reject any value where a term of the original equation is undefined, like tan theta at pi over two. We will need this in example two. Got it?

4:45Example 1: sin x = root 3/2, cos x = -1/2, tan x = -1/root 3

Now example one, three quick ones. Solve sine x equals root three over two, cos x equals minus one half, and tan x equals minus one over root three. Find alpha first, then use the formula. Pause the video and try it first. Done? Okay, let's check. Sine pi over three is root three over two, so alpha is pi over three. Then x equals n pi, plus minus one to the power n, times pi over three. Okay? Now cos x equals minus one half. There is a minus sign, so alpha is minus pi over three, right? Oh no, careful, that is wrong! Cos is even, so cos of minus pi over three is plus one half. For cos, alpha lies from zero to pi, and cos of two pi over three is minus one half. So x equals two n pi, plus or minus two pi over three. With me? Finally, tan of minus pi over six is minus one over root three. So alpha is minus pi over six, and x equals n pi minus pi over six. Clear?

5:55Example 2: solve tan 2 theta times tan theta = 1

Now a harder one. Solve tan two theta, times tan theta, equals one. Remember the third trap. Pause the video and try it first. Done? Okay, let's check. The product is one, so tan theta is not zero, and tan two theta equals one over tan theta. That is cot theta, which equals tan of pi over two minus theta. Right? Both sides are tan now, so two theta equals n pi, plus pi over two minus theta. So three theta equals n pi plus pi over two, and theta equals n pi over three, plus pi over six; the double arrow means implies. Okay? So, is this the final answer? Try n equals one before you write it down. At n equals one, pi over three plus pi over six gives pi over two, where tan theta does not exist! The same happens at n equals four, seven, every third n; the three dots mean and so on. These odd multiples of pi over two are rejected. With me? Look at what is left. The values n equals zero, two, three and five give pi over six, five pi over six, seven pi over six and eleven pi over six. Each one is a multiple of pi, plus or minus pi over six. Got it? So theta equals n pi plus or minus pi over six. Quick check at pi over six: tan pi over three is root three, tan pi over six is one over root three, and their product is one. Clear?

7:27Quick revision: sin, cos, tan formulas and three traps

So, quick revision. Sine gives n pi, plus minus one to the power n, times alpha; cos gives two n pi plus or minus alpha, and tan gives n pi plus alpha, with n an integer. Squares give n pi plus or minus alpha, and pick alpha in the right interval. And watch the traps: cancelling a factor that can be zero, squaring, and undefined values. Now try the practice problems on this page. All the best!