Let S be the sum of all solutions (in radians) of the equation \(\sin^4\theta + \cos^4\theta - \sin\theta \cos\theta = 0\) in [0, 4\(\pi\)]. Then \(\frac{8S}{\pi}\) is equal to _________.
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When solving trigonometric equations, the first step is often to simplify the expression to involve a single trigonometric function and a single angle, as was done here by converting everything to \(\sin(2\theta)\). Once you have a basic equation like \(\sin(u)=k\), find the principal solutions and then use the periodicity of the function to find all solutions within the given interval.
Step 1: Simplify the trigonometric equation.
We use the identity \(\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta\).
Also, we use the double angle identity \(\sin(2\theta) = 2\sin\theta\cos\theta\), which means \(\sin\theta\cos\theta = \frac{1}{2}\sin(2\theta)\).
Substituting these into the equation:
\[ (1 - 2\sin^2\theta\cos^2\theta) - \sin\theta\cos\theta = 0 \]
\[ 1 - 2\left(\frac{1}{2}\sin(2\theta)\right)^2 - \frac{1}{2}\sin(2\theta) = 0 \]
\[ 1 - 2\left(\frac{1}{4}\sin^2(2\theta)\right) - \frac{1}{2}\sin(2\theta) = 0 \]
\[ 1 - \frac{1}{2}\sin^2(2\theta) - \frac{1}{2}\sin(2\theta) = 0 \]
Multiply the entire equation by -2 to clear the fractions and make the squared term positive:
\[ \sin^2(2\theta) + \sin(2\theta) - 2 = 0 \]
Step 2: Solve the quadratic equation for \(\sin(2\theta)\).
Let \(y = \sin(2\theta)\). The equation is \(y^2 + y - 2 = 0\).
Factoring the quadratic gives:
\[ (y+2)(y-1) = 0 \]
This gives two possible solutions for y: \(y = -2\) or \(y = 1\).
- Since the range of the sine function is [-1, 1], \(\sin(2\theta) = -2\) is not possible.
- Therefore, the only valid solution is \(\sin(2\theta) = 1\).
Step 3: Find all solutions for \(\theta\) in the interval [0, 4\(\pi\)].
Let \(u = 2\theta\). The interval for \(\theta\) is \([0, 4\pi]\), so the interval for u is \([0, 8\pi]\).
We need to solve \(\sin(u) = 1\) for \(u \in [0, 8\pi]\).
The general solution for \(\sin(u) = 1\) is \(u = 2n\pi + \frac{\pi}{2}\), where n is an integer.
We find the values of n that give u in the interval \([0, 8\pi]\):
- n=0: \(u = \pi/2\)
- n=1: \(u = 2\pi + \pi/2 = 5\pi/2\)
- n=2: \(u = 4\pi + \pi/2 = 9\pi/2\)
- n=3: \(u = 6\pi + \pi/2 = 13\pi/2\)
(For n=4, \(u = 8\pi + \pi/2\), which is outside the interval).
Now, find the corresponding values of \(\theta = u/2\):
\[ \theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4} \]
Step 4: Calculate the sum S and the final value.
The sum S of all solutions is:
\[ S = \frac{\pi}{4} + \frac{5\pi}{4} + \frac{9\pi}{4} + \frac{13\pi}{4} = \frac{(1+5+9+13)\pi}{4} = \frac{28\pi}{4} = 7\pi \]
The required value is \(\frac{8S}{\pi}\):
\[ \frac{8S}{\pi} = \frac{8(7\pi)}{\pi} = 56 \]
