Back to Problem List. Therefore, the solutions to the first equation will yield the only solutions to our original equation. Also, in the above example we put in a little more explanation than we’ll show in the remaining examples in this section to remind you how these work. Not all equations in this form can be easily solved, however some can so we want to do a quick example of one. So, in this case we get a single solution of 0.7743. Okay, the first step here is identical to the problems in the previous section. To finish the solution here we’ll simply recall the definition of secant in terms of cosine and convert this into an equation involving cosine instead and we already know how to solve those kinds of trig equations. The second angle where we will get the same value of tangent will be exactly opposite of the given point. In this case we can simply divide both sides by a cosine and we’ll get a single tangent in the equation.
So, if we can remember these rules we will be able to determine the remaining angle in \(\left[ {0,2\pi } \right]\) that also works for each solution. Because this document is also being prepared for viewing on the web we split this section into two parts to keep the size of the pages to a minimum. We can now see that this really is an equation that doesn’t involve a sine or a cosine. If we’d been given an interval it would be easy enough to find the solutions that actually fall in the interval. This example is designed to remind you of certain properties about sine and cosine.
The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies. Also, we are going to use 4 decimal places of accuracy in the work here. Searching for the missing side or angle in a right triangle, using trigonometry? Now, we need to properly deal with the 3, so divide that out to get all the solutions to the trig equation. We now have a product of two terms that is zero and so we know that we must have.
First, let’s acknowledge that the values of 3\(t\) that we need are.
Next, solving \(2 + 11\sin \left( {3y} \right) = 0\) gives. For the second angle we’ll note that the lines in the third and fourth quadrant make an angle of 0.7297 with the \(x\)-axis. Hint : Find all the solutions to the equation without regard to the given interval.
Hint : Find all the solutions to the equation without regard to the given interval.
Hint : Find all the solutions to the equation without regard to the given interval. Again, it can’t be stressed enough that while calculators are a great tool if we don’t understand how to correctly interpret/use the result we can (and often will) get the solution wrong. The transformation can be done by using different Now, as mentioned prior to starting the example this quadratic does not factor. Type 2-3 given values in the second part of the calculator and in a blink of an eye you'll find the answer.
First notice that, in this case, if we plug in negative values of \(n\) we will get negative solutions and these will not be in the interval and so there is no reason to even try these. In the quotient however, the difference in signs will cancel out and we’ll get the same value of tangent. Using our calculator we can see that. Most calculators these days will have buttons on them for these three so make sure that yours does as well. Before proceeding with the problems we need to go over how our calculators work so that we can get the correct answers. We now need to move into a different type of trig equation. You need only two given values in the case of: Remember that if you know two angles, it's not enough to find the sides of the triangle. There really isn’t too much to do with this problem. So, to solve this equation we’ll first get all the terms on one side of the equation and then factor an \(x\) out of the equation. This trigonometry calculator will help you in two popular cases when trigonometry is needed. Since the line segment in the first quadrant forms an angle of 0.7227 radians with the positive \(x\)-axis then so does the line segment in the fourth quadrant.
We weren’t given an interval in this problem so here is nothing else to do here. As with the previous example we’ll use the positive choice, but that is purely a matter of preference. Having our calculator compute \({\cos ^{ - 1}}\left( {{\frac{3}{4}}} \right)\) and hence solve \(\cos \left( x \right) = {\frac{3}{4}}\) gives. If has degree , then it is well known that there are roots, once one takes into account multiplicity.