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Chapter 5: Problem 38

Explain why the equation $$ (\sin x)^{2}-4 \sin x+4=0 $$ has no solutions.

Short Answer

Expert verified
The given equation \((\sin x)^2 -4\sin x + 4 = 0\) leads to a unique quadratic solution \(\sin x = 2\). However, as the sine function has a range of \([-1, 1]\), the obtained value is not valid and hence, the equation has no solutions.

Step by step solution

01

Write the equation in quadratic form

The given equation can be written in the form of a quadratic equation regarding the sine function: \[ (\sin x)^2 -4\sin x + 4 = 0 \]
02

Check the range of sine function

We know that the sine function is bounded in the interval \([-1, 1]\): \[ -1 \leq \sin x \leq 1 \]
03

Apply the discriminant method to the quadratic equation

The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by: \[ \Delta = b^2 - 4ac \] Let's find the discriminant (\(\Delta\)) of our trigonometric equation: \[ \Delta = (-4)^2 - 4 \cdot 1 \cdot 4 = 16 - 16 = 0 \] The discriminant is equal to zero, which means the quadratic equation has a unique real solution. Let's find that solution:
04

Solve the quadratic equation using the discriminant value

Since the discriminant is equal to zero, the solution of the quadratic equation is given by: \[ \sin x = \frac{-b}{2a} = \frac{4}{2\cdot1} = 2 \]
05

Analyze the obtained solution

From our analysis in Step 2, we know that the sine function has a range of \([-1, 1]\). The solution we have found, \(\sin x = 2\), is outside that range. Therefore, this value is not valid for the sine function.
06

Conclude the solution

Since the unique solution of the equation \((\sin x)^2 -4\sin x + 4 = 0\) is outside the range of possible values for the sine function, we can conclude that the equation has no solutions.

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