cos/sin use 2npi and tan uses npi, but how do you know when to add 2npi or npi to the solution of the equation?
ive had a problem where it goes cos pi/8 + npi and cos 2npi - 7pi/8. then you plug in for n to get the solutions. I'm not understanding where to put 2npi or npi it's not a period interval cause sin and cos are usually 2npi. Very confused when to put it or not put it.
When solving trigonometric equations, the values for cosine (cos), sine (sin), and tangent (tan) repeat themselves with a certain pattern. Adding multiples of 2π or π to the values results in the same trigonometric function value.
To determine whether to add 2nπ or nπ, you need to consider the specific properties of the equation you're solving and the restrictions of the domain.
1. Cosine (cos):
- The general form of the equation is cos(x) = a, where a is a real number.
- To find the solutions, you can apply the inverse cosine function (arccos) to both sides of the equation.
- The inverse cosine function gives a unique solution in the range [0,π]. However, cos(x) has a period of 2π, meaning it repeats itself every 2π units.
- Therefore, to find all solutions, you need to add 2nπ, where n is an integer, to the initial solution obtained from the arccos function.
Example: If cos(x) = 0, the solution is x = arccos(0) = π/2. However, there are infinitely many solutions. Thus, the complete solution set is x = π/2 + 2nπ.
2. Sine (sin):
- The process for finding solutions for sine is similar to cosine.
- The general form of the equation is sin(x) = a, where a is a real number.
- Applying the inverse sine function (arcsin) gives a unique solution in the range [-π/2,π/2].
- Since sin(x) also has a period of 2π, you add 2nπ to the initial solution to obtain all possible solutions.
Example: If sin(x) = 1/2, the solution is x = arcsin(1/2) = π/6. But the complete solution set is x = π/6 + 2nπ.
3. Tangent (tan):
- The general form of the equation is tan(x) = a, where a is a real number.
- Applying the inverse tangent function (arctan) gives a unique solution in the range (-π/2,π/2).
- Tangent function does not have a period of 2π like sine and cosine.
- To find all solutions, you add nπ to the initial solution obtained from arctan.
Example: If tan(x) = √3, the solution is x = arctan(√3) = π/3. The complete solution set is x = π/3 + nπ.
In summary, the decision to add 2nπ or nπ when obtaining solutions for trigonometric equations depends on the specific function being used (cos, sin, or tan) and the properties of the equation being solved.
tan(kx) has period pi/k, since x grows k times faster.
So, I think you have a typo. It should be
3tan3x = 3
tan3x=1
now, you know that tan pi/4 = 1, so
3x = pi/4, and x = pi/12. Now, tan(3x) has period pi/3.
2tanx-2 = 0
tanx = 1
Since tan pi/4 = 1, tan (pi/4 + pi) = tan 5pi/4 =1 as well.
The 2nd has solutions at pi/4 + n*pi for any n. If they chose 5pi/4, there must have bee some conditions on the domain.
okay,
thanks!
I have an equation that is 3tan2x=3 and the solution is x=pi/12 + npi/3
and another equation that is 2tanx-2=0 , the answer sheet says the solution is x=5pi/4.
why is the npi not added to the second solution but it is on the first?
as you said, sin and cos have period of 2pi
and tan has period pi.
That's how you know.