Video: Solving Trigonometric Equations Involving Special Angles

Find the smallest positive angle that satisfies both 2 cos ๐œƒ โˆ’ โˆš(2) = 0 and tan ๐œƒ โˆ’ 1 = 0.

03:02

Video Transcript

Find the smallest positive angle that satisfies both two cos ๐œƒ minus root two equals zero and tan ๐œƒ minus one equals zero.

Weโ€™re going to begin by solving each of these equations. Thatโ€™s two cos ๐œƒ minus root two equals zero and tan ๐œƒ minus one equals zero. Weโ€™re solving for ๐œƒ, so the first thing weโ€™re going to do with our first equation is add the square root of two to both sides. That leaves us with simply two cos ๐œƒ on the left-hand side and root two on the right. Next, we divide both sides of our equation by two. And so we get cos of ๐œƒ is equal to root two over two. Now, at this point, what we could do next is find the inverse or arccos of both sides of our equation. Weโ€™d have ๐œƒ is equal to the inverse cos of root two over two.

However, this is one of the values we should know by heart. We know that cos of 45 degrees is equal to root two over two. So here, for cos of ๐œƒ to be equal to root two over two, ๐œƒ must be equal to 45 degrees. Now, in fact, if we think about the cosine function, we know that itโ€™s periodic. But itโ€™s also symmetrical about the line ๐‘ฅ equals 180. And so in the interval zero to 360 degrees, thereโ€™s another solution. And that solution is 360 minus 45 which is 315 degrees. Now since the cosine function, as we said, is periodic and it repeats every 360 degrees, we can find further solutions by adding multiples of 360 degrees to both of these solutions. However, weโ€™re trying to find the smallest positive angle that satisfies both of our equations, so this might be sufficient.

Letโ€™s move on to the second equation. This one requires distinctly fewer steps. Weโ€™re going to add one to both sides of this equation. And that gives us tan of ๐œƒ is equal to one. Once again, we could then find the inverse tan of both sides of our equation so that ๐œƒ is equal to the inverse tan of one. However, this is also one of the special angles that we should know. tan of 45 is equal to one, so ๐œƒ must in fact be equal to 45 degrees. And so we see we didnโ€™t actually need to worry about finding any further solutions. We know that ๐œƒ is equal to 45 degrees satisfies both of these equations.

We can convince ourselves that this is indeed the smallest positive angle that satisfies both of our equations by thinking of the shape of the tan graph for values of ๐œƒ greater than zero. We have one solution of ๐œƒ equals 45 degrees, and then the other solution within the interval 0 to 360 at least is indeed bigger than this. Itโ€™s 180 plus 45, so thatโ€™s 225 degrees. And so the smallest positive angle that satisfies both of our equations is 45 degrees.

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