Solve each equation for exact solutions over the interval [0, ...

Question

Solve each equation for exact solutions over the interval [0, 2π).

―2 sin² x = 3 sin x + 1

Accepted Answer

Hey, everyone here we are asked to find the solutions of the given equation over the interval from open bracket 022 pi close parentheses. Here we are given the equation negative two multiplied by cosine squared of X is equal to three, multiplied by sine of X minus three. We have four answer choice options. Answer A, the solution set of pi divided by six pi divided by two and five pi divided by six. Answer B, the solution set of pi divided by six and pi divided by two answer C the solution set of pi divided by six and five pi divided by six and answer D no solution. So before we can begin finding our solutions for the given equation, we first want to simplify this equation so that we just have one trigonometric function. So here, since we have this cosine squared of X, we just want to recall the Pythagorean identity which states that sine squared of X plus cosine squared of X is equal to one. And from this formula, we can conclude that cosine squared of X is equivalent to one minus sine squared of X. And so now utilizing the rearranged formula, we can substitute in a new expression for cosine squared of X. And so our new equation becomes negative two multiplied by the quantity of one minus sine squared of X is equal to three multiplied by sine of X minus three. And now continuing to simplify, we just want to distribute in this negative two. And so we now have negative two plus two multiplied by sine squared of X is equal to three multiplied by sine of X minus three. And now we just want to rearrange our equation so that all of our terms are on the same side. So rewriting, we have two multiplied by sine squared of X minus three, multiplied by sine of X plus one is equal to zero. And now we just want to factor the left hand side. So factoring we get the quantity of two multiplied by sin of X minus one multiplied by the quantity of sine of X minus one is equal to zero. So now to make our current equation easier to find solutions of, we can recall the zero factor property which allows us to rewrite this equation into two individual equations. And so from the property we know that two multiplied by sin of X minus one can be equal to zero or sine of X minus one is equal to zero. So now we just want to solve both of these individual equations for sin of X. So beginning with two multiplied by sin of X minus one is equal to zero. We need to solve for sin of X by first adding one to both sides. And so we have two multiplied by sine of X is equal to one. Now we just need to divide both sides by the coefficient which is two. So we find that sine of X is equal to one half. Now moving back to our second equation where we have sin of X minus one is equal to zero, we want to solve four sin of X by simply adding one to both sides. And so here we have that sine of X is equal to positive one. So now our last step is to recall our unit circle so that we can find our solutions to these final two simplified equations. So beginning with sin of X is equal to one half. Well, from the unit circle, we find that the X value that makes makes this equation true is pi divided by six and five pi divided by six. So so far we have found two possible X values moving on to our second equation. We need to find the X value which causes sin of X to be equivalent to one. And from the unit circle, we find that this X value is pi divided by two. So now we have found three different solutions for our given equation. So we can just simplify our findings by rewriting them as a solution. Set. And so we have the solution set of pi divided by six pi divided by two and five pi divided by six. And so with this final solution set, we are left here with answer choice. A where again, we have the solution set of pi divided by six pi divided by two and five pi divided by six. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Solve each equation for exact solutions over the interval [0, 2π).
―2 sin² x = 3 sin x + 1

Key Concepts

Trigonometric Equations

Quadratic Form in Trigonometric Functions

Interval and Exact Solutions in [0, 2π)

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