trig Identities 13

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Solve the following equation for : .0<x<2pie:sin2x-sinx-2= Select all that apply.

Dec 17th, 2014


Solve the following equation for : .0<x<2pie:sin2x-sinx-2= Select all that apply.

sin^2 x - sin x - 2 = 0 for 0 < x < 360
we might consider using the Trig identity sin^2(x) = 1 - cos^2(x) to substitute for sin^2(x). But it doesn't seem to "improve" the equation. It doesn't look like the new equation would be easier to solve.
If we look at this from an algebraic viewpoint we will see a solvable quadratic equation. It might be easier to see this if we use a temporary variable. Let s = sin(x), Then our equation is

s^2 - s - 2 = 0
This is an easily solved quadratic equation. Just factor it:
(s-2)(s+1) = 0
Set the factors equal to zero and solve:
s-2 = 0 or s+1 = 0
s = 2 or s = -1
Of course we are not looking for "s". We are looking for x. So we substitute back for "s" ...
sin(x) = 2 or sin(x) = -1
(Eventually you will learn to "see" the quadratic nature of your original equation and factor and solve it without the use of a temporary variable.)

Now we can use our Trig. finish the solution:
1) sin(x) = 2. When is the sin of an angle 2? Answer: Never!! This means that there is no solution possible fron this equation.
2) sin(x) = -1. When is the sin of an angle -1? Answer: Any angle which is coterminal with 270. So the general solution is:
x = 270 + 360n
But we are asked only to find solutions between 0 and 360 and there is only one angle between 0 and 360 whose sin is -1: 270.

Dec 17th, 2014

Dec 17th, 2014
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