Question

Find the general solution for the equation \(4\left(16^{\cos ^{2} x}\right)=2^{6 \cos x} .\) Give your answer in radians.

Short answer

The general solutions are \(x = 2k\pi\), \(x = \frac{\pi}{3} + 2k\pi\), and \(x = -\frac{\pi}{3} + 2k\pi\).

Step-by-step solution

Rewrite the Equation in Base 2

Start by expressing both sides of the equation with the same base, base 2. We know that \[16 = 2^4,\] so we can rewrite the left side as \[4(16^{\text{cos}^2 x}) = 4((2^4)^{\text{cos}^2 x}) = 4(2^{4\text{cos}^2 x}).\] Now the equation becomes: \[4(2^{4\text{cos}^2 x}) = 2^{6\text{cos} x}.\]

Simplify the Multiplication

The next step is to combine the multiplication on the left side. Note that \[4 = 2^2,\] so the left side can be rewritten as: \[2^2 \times 2^{4\text{cos}^2 x} = 2^{2 + 4\text{cos}^2 x}.\] Thus, the equation becomes:\[2^{2 + 4\text{cos}^2 x} = 2^{6\text{cos} x}.\]

Set the Exponents Equal

Since the bases are the same, set the exponents equal to each other:\[2 + 4\text{cos}^2 x = 6\text{cos} x.\]

Form a Quadratic Equation

Rearrange the equation to form a standard quadratic equation:\[4\text{cos}^2 x - 6\text{cos} x + 2 = 0.\]

Use the Quadratic Formula

Use the quadratic formula \[\text{cos} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] where \(a = 4\), \(b = -6\), and \(c = 2\). Substitute these values into the formula:\[\text{cos} x = \frac{6 \pm \sqrt{(-6)^2 - 4(4)(2)}}{2(4)} = \frac{6 \pm \sqrt{36 - 32}}{8} = \frac{6 \pm \sqrt{4}}{8} = \frac{6 \pm 2}{8}.\]

Solve for cos x

Solve the above equation for the two values:\[\text{cos} x = \frac{6 + 2}{8} = 1,\] and \[\text{cos} x = \frac{6 - 2}{8} = \frac{1}{2}.\]

Find x in Radians

For \(\text{cos} x = 1\), the solution in radians is \(x = 2k\pi\) where \(k\) is an integer.For \(\text{cos} x = \frac{1}{2}\), the general solutions in radians are \(x = \frac{\pi}{3} + 2k\pi\) and \(x = -\frac{\pi}{3} + 2k\pi\) where \(k\) is an integer.

Key concepts

Cosine Function

The cosine function is one of the basic trigonometric functions used in mathematics. It is defined as the ratio of the adjacent side to the hypotenuse in a right-angled triangle. The function has a range from -1 to 1, meaning that for any angle, the cosine value will always lie within this interval.
The cosine function is periodic, with a period of \(2\pi\). This means that the function repeats its values every \(2\pi\) radians.
Cosine is often used in solving trigonometric equations, such as the given problem where it is used to express the equation in a more manageable form. Understanding the properties of the cosine function is key to solving such equations. Its graph is symmetric about the y-axis, which is another useful feature when solving trigonometric problems.

Quadratic Formula

The quadratic formula is a universal method for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula provides the solutions to the quadratic equation by using the coefficients \(a\), \(b\), and \(c\).
In the context of our problem, after setting the exponents equal to each other and rearranging terms, we form a quadratic equation in terms of \(\cos x\):
\[ 4\cos^2 x - 6\cos x + 2 = 0 \] Substituting \(a = 4\), \(b = -6\), and \(c = 2\) into the quadratic formula, we solve for \(\cos x\), finding two possible solutions:
\[ \cos x = \frac{6 \pm \sqrt{36 - 32}}{8} = \frac{6 \pm 2}{8} \] This leads to \(\cos x = 1\) and \(\cos x = \frac{1}{2}\).

Solutions in Radians

In trigonometry, solutions are often given in radians rather than degrees. Radians are a measure of angle based on the radius of a circle. One full circle is \(2\pi\) radians, which is equivalent to 360 degrees.
For the solutions of our problem, we need to express them in radians. For \(\cos x = 1\), the angle \(x\) can be \(0\), \(2\pi\), \(4\pi\), etc., or generally, \(x = 2k\pi\) where \(k\) is an integer.
For \(\cos x = \frac{1}{2}\), the angles \(\frac{\pi}{3}\) and \(\frac{-\pi}{3}\) are solutions within one period \(0\) to \(2\pi\). However, because the cosine function is periodic with a period of \(2\pi\), these solutions are repeated every \(2\pi\). Thus, the general solution in radians for this case is \(x = \frac{\pi}{3} + 2k\pi\) and \(x = -\frac{\pi}{3} + 2k\pi\) where \(k\) is an integer.

General Solution

The general solution to a trigonometric equation provides all possible values of the variable that satisfy the equation. This is particularly important for periodic functions like cosine, where multiple solutions can exist within one period.
For our problem, we found specific solutions \(\cos x = 1\) and \(\cos x = \frac{1}{2}\).
Since \(\cos x = 1\) corresponds to the angles where the cosine curve reaches its maximum, the general solution is \(x = 2k\pi\) where \(k\) is an integer.
Similarly, for \(\cos x = \frac{1}{2}\), we account for all the repeated cycles of the cosine function. Therefore, the general solutions are \(x = \frac{\pi}{3} + 2k\pi\) and \(x = -\frac{\pi}{3} + 2k\pi\) where \(k\) is an integer.
Combining these, the general solution to the original equation \( 4(16^{\cos^2 x}) = 2^{6\cos x} \) in radians is:
\[ x = 2k\pi \quad \text{or} \quad x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad x = -\frac{\pi}{3} + 2k\pi \] where \(k\) is any integer.