Video Transcript
Find the solution set of 𝜃 that
satisfies root three multiplied by the csc of 90 degrees minus 𝜃 minus two equals
zero, where 𝜃 lies on the closed interval from zero degrees to 180 degrees.
In this question, we begin by
considering the reciprocal trigonometric functions. We know that the csc of any angle
𝛼 is equal to one over the sin of angle 𝛼. Before using this identity, we will
let 𝛼 equal 90 degrees minus 𝜃. This enables us to solve the
simpler equation root three csc of 𝛼 minus two equals zero. Adding two to both sides of this
equation and then dividing through by root three, we have the csc of 𝛼 equals two
over root three. Using the reciprocal identity, we
have sin 𝛼 is equal to one over two over root three, which is equal to root three
over two.
From our knowledge of special
angles, we know that the sin of 60 degrees is equal to root three over two. This means that a solution to the
equation sin 𝛼 equals root three over two is 𝛼 equals 60 degrees. We are looking for solutions of 𝜃
between zero and 180 degrees inclusive. To write this interval in terms of
𝛼, we subtract 90 degrees from the inequality such that 𝛼 is greater than or equal
to negative 90 degrees and less than or equal to 90 degrees. Using a CAST diagram, we can see
we’re only looking for solutions in the first or fourth quadrants. As the sin of angle 𝛼 is positive,
there will therefore only be a solution in the first quadrant. This is the solution we have
already found: 𝛼 is equal to 60 degrees.
As 𝛼 is equal to 90 degrees minus
𝜃, then 𝜃 must be equal to 90 degrees minus 𝛼. Substituting in our value of 𝛼, we
have 𝜃 is equal to 90 degrees minus 60 degrees, which gives us an answer of 30
degrees. The solution set that satisfies the
equation root three multiplied by the csc of 90 degrees minus 𝜃 minus two equals
zero is 30 degrees.
