Solve each equation over the interval [0°, 360°). Write soluti...

Question

Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth, as appropriate.

csc² θ ―2 cot θ = 0

Accepted Answer

Hey, everyone here we are asked to find the solutions of the given equation over the interval from open bracket zero degrees to 360 degrees. Close parentheses. Here we are given the equation second squared of theta minus two multiplied by tangent of theta is equal to zero. Here we have four answer choice options. Answer A, the solution set of 45 degrees and 225 degrees. B the solution set of 225 degrees. See the solution set of 45 degrees and answer D no solution. So to begin this problem, we first want to start by simplifying our given equation so that we are only dealing with one trigonometric function. So we can do this by recalling the trigonometric identity for second squared of theta. And this identity tells us that second squared of theta is equivalent to one plus tangent squared of theta. So now we can substitute in our new expression for a sequence squared of theta into our given equation. And so we now have the equation one plus tangent squared of theta minus two multiplied by tangent of theta is equal to zero And now we just want to manipulate our current equation. So we can solve four tangent of the. And so the first thing we can notice is that the left hand side of our current equation matches the expanded form of the difference of squares formula. So now just recalling the difference of squares formula, we know that the quantity of A minus B squared is equivalent to A squared minus two multiplied by a multiplied by B plus B squared. And so now we can notice that the right hand side of the formula matches the format of the left hand side of our current equation. And so we can identify that our value for A is tangent of theta and our value for B is just one. So now we can condense our expanded left hand side by utilizing the left hand side of the formula. And so rewriting our current equation, we now have the quantity of tangent of theta minus one squared is equal to zero. And now that we have rewritten our equation, it is much more easier to solve this equation for a tangent of theta. So all we need to do right now is take the square root of both sides. So we get rid of the exponent. And so doing this, we now have tangent of theta minus one is equal to zero. And now our last step in solving for a tangent of theta is to just add one to both sides. And so we find that tangent of theta is equal to one. And now, in order to find our solutions for our equation, we need to find the theta values that cause tangent of theta to be equivalent to one. So for this, we need to recall the unit circle and to help us identify our theta values on the unit circle, we can recall that tangent of theta is equivalent to sine of theta divided by cosine of theta. And so now identifying our theta values on the unit circle, we find that tangent of theta is equal to one for when theta is equal to 45 degrees and 225 degrees. And with this, we have found two different solutions for our given equation. And so we can summarize our findings by writing these values as a solution set. And so we are left with the solution set of 45 degrees and 225 degrees. And with these two final solutions, we are left here with answer choice. A where again, we have the solution set of 45 degrees and 225 degrees. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth, as appropriate.
csc² θ ―2 cot θ = 0

Key Concepts

Reciprocal and Quotient Identities

Pythagorean Identity involving Cotangent and Cosecant

Solving Trigonometric Equations within a Given Interval

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