Video Transcript
Find the set of values satisfying
the tan of two 𝑥 plus 𝜋 over five is equal to negative one, where 𝑥 is greater
than or equal to zero and less than or equal to two 𝜋.
To solve this equation, we’ll begin
by redefining the argument as this will allow us to use the symmetry of the tangent
function. We will let 𝜃 equal two 𝑥 plus 𝜋
over five. This means that we need to solve
the tan of 𝜃 equals negative one where 𝜃 is greater than or equal to 𝜋 over five
and less than or equal to 21𝜋 over five as we multiply each part of the inequality
by two and then add 𝜋 over five. Next, we recall that for 𝜃
measured in radians, the exact values of tan 𝜃 are as shown. We see that tan of 𝜋 over four is
equal to one. Next, we will sketch the graph of
𝑦 equals the tan of 𝜃. We will then add the horizontal
lines where 𝑦 equals one and 𝑦 equals negative one.
Due to the rotational symmetry of
the tangent function, the first solution occurs when 𝜃 is equal to 𝜋 minus 𝜋 over
four. This is equal to three 𝜋 over
four. As the function is periodic with a
period of 𝜋 radians, we can find the remaining solutions by adding multiples of 𝜋
to this value. Firstly, three 𝜋 over four plus 𝜋
is equal to seven 𝜋 over four. We also have solutions 11𝜋 over
four and 15𝜋 over four. These are the four points of
intersection shown on the graph. Clearing some space and rewriting
our four solutions for 𝜃, we can now calculate the values of 𝑥. As 𝜃 is equal to two 𝑥 plus 𝜋
over five, two 𝑥 is equal to 𝜃 minus 𝜋 over five. Dividing through by two, we have 𝑥
is equal to 𝜃 over two minus 𝜋 over 10.
We can now substitute each of our
values of 𝜃 into this equation. This gives us four values of 𝑥
equal to 11𝜋 over 40, 31𝜋 over 40, 51𝜋 over 40, and 71𝜋 over 40. This is the set of values that
satisfies the equation tan of two 𝑥 plus 𝜋 over five equals negative one where 𝑥
lies between zero and two 𝜋 inclusive.
