Solve the trig equation exactly over the indicated interval.
tanθ = 0, all real numbers
I know the answer is πn, but I don't understand how they got this. I get that tanθ = 0 on π and 2π on the unit circle, so did they just put n on the end of π because of this?
tanØ = sinØ/cosØ = 0
in sinØ/cosØ = 0 , that is only possible if sinØ = 0
and you should know that the basic sine curve has zeros at 0 , π and 2π
since the period of sinØ = π,
a new answer can be obtained by simply adding/subtracting multiples of π (or nπ) to any anwer
So why not start with 0 + nπ = nπ
Use trigonometric identities to solve tan(2θ)+tan(θ)=0 exactly for 0≤θ≤π. If there is more than one answer, enter your answers as a comma separated list
To solve the trigonometric equation tanθ = 0 for all real numbers, we need to find all values of θ that satisfy this equation.
First, let's recall the definition of the tangent function. The tangent of an angle θ is defined as the ratio of the sine of θ to the cosine of θ, which is given by tanθ = sinθ/cosθ. Since we want to find θ when the tangent is equal to 0, we set the numerator, sinθ, equal to 0, which means sinθ = 0.
The sine of an angle is equal to 0 at two specific angles: 0 and π. So, we have sinθ = 0 when θ = 0 and θ = π.
Now, let's take a closer look at the tangent function. On the unit circle, the tangent is positive in the first and third quadrants, and negative in the second and fourth quadrants. The tangent is equal to 0 at θ = 0 on the unit circle.
Since the tangent function has a period of π (or 180 degrees), for every solution θ = 0, we can add any multiple of π to it and get another solution. Hence, the general solution to the equation tanθ = 0 for all real numbers can be written as:
θ = 0 + nπ,
where n is an integer. This general form allows us to include all possible solutions to the equation.
Therefore, the answer to the equation tanθ = 0 for all real numbers is θ = πn, where n represents any integer.