Standard Methods

Transcript

0:00Introduction

Hello everyone! Let us start with a small mystery. Take the equation three cos x plus four sine x equals minus three. One clever substitution turns it into plain algebra, and gives one family of answers. But now put x equal to pi. Three cos pi is minus three, and four sine pi is zero. So pi is also a root, and the method never found it! Today we learn the standard methods, and find out exactly where that root went. These come up often in JEE, so let's begin.

0:34Factorise, or make a quadratic

So, the first method is the simplest. If the equation can be factorised, factorise it, often after the sum-to-product or multiple-angle formulas. Each factor equal to zero then gives a basic equation. Okay? Next, if the equation mixes, say, sine squared x and cos x, change to one ratio. Use sine squared x equals one minus cos squared x; the small two after sine means squared. Now it is just a quadratic in cos x. Right? But are both roots of that quadratic always allowed? Let me ask you this: which values can cos x actually take? Only values from minus one to one, yes, exactly! The sign with a line under it means less than or equal to. So keep only the roots in the range of that ratio, and reject the rest.

1:27The form a cos theta + b sin theta = c

Clear? Now, the most important form: small a cos theta, plus small b sine theta, equals small c. Theta, the Greek letter, is the unknown angle. Divide by r, the square root of a squared plus b squared; the tick-shaped sign is the square root. Then small a over r and small b over r are the cos and the sine of a fixed angle, phi, another Greek letter. With me? So the left side becomes cos theta cos phi, plus sine theta sine phi, which is just cos of theta minus phi. So cos of theta minus phi equals small c over r. Okay? Now tell me, when does this equation have a solution at all? Only when small c over r lies between minus one and one, yes! So solutions exist if and only if mod c is at most the root of a squared plus b squared; the two vertical bars mean mod c, its size without the sign. Got it?

2:22Half-angle substitution t = tan(theta/2)

Next, a substitution for any mix of sine theta and cos theta: put t equal to tan of theta over two. Then sine theta is two times t, over one plus t squared, and cos theta is one minus t squared, over one plus t squared. Both are rational in t, so it is plain algebra now. Nice, isn't it? But tan is not defined everywhere. So where is tan of theta over two undefined? Tan is undefined where cos is zero, so when theta over two is an odd multiple of pi over two. That means theta equals two n plus one, times pi, with n any integer; here pi radians means one hundred and eighty degrees. Clear? And this is the big trap. At those angles there is no t, so the substitution cannot see them, and every such root is lost. So always test theta equals two n plus one, times pi, in the original equation. That is where the root from our mystery went. With me?

3:22Two more substitutions

When sine x plus cos x comes with sine x times cos x, call the sum t. Squaring, sine x cos x is t squared minus one, over two. For sine x minus cos x, only the sign flips; the sign with plus above minus means plus or minus. And mod t is at most root two, by the form we just learnt. Okay? And sometimes the equation comes dressed up with powers, like a number small a, raised to the power sine squared x. Put u equal to that, solve for u first, then come back to x. Clear?

3:56Example 1: solve root 3 cos x + sin x = root 2

Now the first example. Solve root three cos x, plus sine x, equals root two. Pause the video and try it first. Done? Okay, let's check. Small a is root three and small b is one, so r is the root of three plus one, which is two. Root two is less than two, so solutions exist. Right? Divide everything by two. Root three over two, times cos x, plus one half times sine x, equals one over root two. Okay? Now, which angle has its cos equal to root three over two, and its sine equal to one half? It is pi over six, yes, exactly! So the left side is cos of x minus pi over six, and one over root two is cos of pi over four. Clear? When two cosines are equal, the angles differ by two n pi, with a plus or minus sign. So x minus pi over six equals two n pi, plus or minus pi over four. With me? Add pi over six to both sides. Pi over six plus pi over four is five pi over twelve, and pi over six minus pi over four is minus pi over twelve. So x equals two n pi plus five pi over twelve, or two n pi minus pi over twelve. The curly e-shaped sign means belongs to, and the double-struck Z is the integers. Got it?

5:21Example 2: 3 cos x + 4 sin x = k, and the lost root x = pi

Now the harder one, our mystery from the start. For which k does three cos x plus four sine x equal k have a real solution? Then solve it for k equal to minus three. Pause the video and try it first. Done? Okay, let's check. Here r is the root of nine plus sixteen, which is five. So a real solution exists if and only if mod k is at most five. Okay? Now k is minus three, and we put t equal to tan of x over two. Substitute, and multiply through by one plus t squared. Three times one minus t squared, plus eight times t, equals minus three times one plus t squared. Right? Expand, and minus three times t squared appears on both sides, so the t squared terms cancel! Eight times t equals minus six, so t is minus three over four. Clear? Tan inverse of three over four is the angle between minus pi over two and pi over two whose tan is three over four; the small minus one up top means inverse, not a power. Since tan of x over two is minus three over four, x over two equals n pi minus tan inverse three over four, and x equals two n pi minus two tan inverse three over four. With me? Is that the complete answer? Tell me, did the substitution find every root? No, it did not, and this is the trap! Test x equals two n plus one, times pi. Three cos pi is minus three, and four sine pi is zero, so the total is minus three, and it works. That is our missing root, found at last. Got it? So the full answer is x equals two n pi minus two tan inverse three over four, or x equals two n plus one, times pi. Miss the second family, and you lose marks. Okay?

7:12Quick revision: when solvable, range check, tan(x/2) trap

So, quick revision. Small a cos theta plus small b sine theta equals small c is solvable if and only if mod c is at most the root of a squared plus b squared. For a quadratic in one ratio, keep only the roots inside its range. And with t equal to tan of theta over two, always test theta equals two n plus one, times pi, separately. Now go and try the practice problems on this page. All the best!