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Chapter 3: Problem 220

If \(0 \leq \mathrm{x} \leq \pi\) and \(16^{(\sin ) 2 \mathrm{x}}+16^{(\cos ) 2 \mathrm{x}}=10\) then \(\mathrm{x}=\) (a) \((\pi / 3)\) (b) \((\pi / 2)\) (c) \((\pi / 4)\) (d) \((3 \pi / 4)\)

Short Answer

Expert verified
The correct answer is (c) \(x=\frac{\pi}{4}\).

Step by step solution

01

Use Trigonometric Identity

We can use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). We need to find a value of x such that \(16^{\sin^2 x} + 16^{\cos^2 x} = 10\). Since both sin and cos can be represented in terms of each other, let's use the given identity to make a substitution. We will substitute for \(\cos^2 x\) by rewriting it as \(\cos^2 x = 1 - \sin^2 x\).
02

Substitute and Simplify

Now we substitute \(\cos^2 x = 1 - \sin^2 x\) in the given equation: \(16^{\sin^2 x} + 16^{1 - \sin^2 x} = 10\). Now let's make another substitution to simplify the equation further. Let \(y = \sin^2 x,\) so we have: \(16^y + 16^{1 - y} = 10\).
03

Use Properties of Exponents

Observe that \(16^{1-y} = \frac{1}{16^y}\). So we can rewrite the equation as: \(16^y + \frac{1}{16^y} = 10\). Now, multiply both sides by \(16^y\) to get rid of the fraction: \((16^y)^2 + 1 = 10 \times 16^y\).
04

Transform the Equation into Quadratic Form

Now let's make a substitution \(z = 16^y\). The equation becomes: \(z^2 + 1 = 10z\). Rearranging the equation into a quadratic equation: \(z^2 - 10z + 1 = 0\).
05

Solve the Quadratic Equation

Now we need to find the roots of the equation: \(z^2 - 10z + 1 = 0\). This is a quadratic equation in the form of \(ax^2 + bx + c = 0\). We can find the roots using the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Plugging in the values, we get: \(z = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(1)}}{2(1)} = \frac{10 \pm \sqrt{96}}{2}\). Since the value of \(16^y\) must be positive, we only consider the positive root: \(z = \frac{10 + \sqrt{96}}{2}\).
06

Back-substitute and Find x

Now we can back-substitute \(z\) and \(y\) to find x: \(16^y = z = \frac{10 + \sqrt{96}}{2}\). Therefore, \(\sin^2 x = y = \log_{16} \frac{10 + \sqrt{96}}{2}\). Taking the square root of both sides, we get: \(\sin x = \sqrt{\log_{16} \frac{10 + \sqrt{96}}{2}}\). As x lies in the range of \(0 \le x \le \pi\), and given the options, evaluating each option, we find that \(\boxed{x=\frac{\pi}{4}}\) (option (c)) satisfies the equation \(16^{\sin^2 x} + 16^{\cos^2 x} = 10\).

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