Question

Solve for all real x x satisfying cos 4 x + cos 2 x sin 2 x + sin 4 x = 3 4 . cos 4 x+cos 2 xsin 2 x+sin 4 x= 4 3 ​ .

Answer 1

x = \frac{n\pi}{2} + \frac{1}{2} \arctan\left(\frac{6 \pm \sqrt{43}}{7}\right), n \in \mathbb{Z}

Answer 2

The given trigonometric equation is: $\cos 4x + \cos 2x \sin 2x + \sin 4x = \frac{3}{4}$ We use the identity $2 \sin \theta \cos \theta = \sin 2\theta$. Thus, $\cos 2x \sin 2x = \frac{1}{2} \sin 4x$. Substituting this into the equation: $\cos 4x + \frac{1}{2} \sin 4x + \sin 4x = \frac{3}{4}$ $\cos 4x + \frac{3}{2} \sin 4x = \frac{3}{4}$ Multiply by 2 to get rid of the fraction: $2 \cos 4x + 3 \sin 4x = \frac{3}{2}$ This is an equation of the form $a \cos \theta + b \sin \theta = c$, where $\theta = 4x$, $a=2$, $b=3$, and $c=\frac{3}{2}$. We use the substitution $t = \tan(\frac{\theta}{2}) = \tan(\frac{4x}{2}) = \tan(2x)$. The identities for $\cos \theta$ and $\sin \theta$ in terms of $t$ are: $\cos \theta = \frac{1 - t^2}{1 + t^2} \quad \text{and} \quad \sin \theta = \frac{2t}{1 + t^2}$ Substituting these into the equation $2 \cos 4x + 3 \sin 4x = \frac{3}{2}$: $2 \left(\frac{1 - t^2}{1 + t^2}\right) + 3 \left(\frac{2t}{1 + t^2}\right) = \frac{3}{2}$ Multiply both sides by $2(1+t^2)$ to clear the denominators: $2 \cdot 2 (1 - t^2) + 2 \cdot 3 (2t) = 3 (1 + t^2)$ $4 (1 - t^2) + 12t = 3 (1 + t^2)$ $4 - 4t^2 + 12t = 3 + 3t^2$ Rearrange the terms to form a quadratic equation in $t$: $7t^2 - 12t - 1 = 0$ We solve this quadratic equation for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(7)(-1)}}{2(7)}$ $t = \frac{12 \pm \sqrt{144 + 28}}{14}$ $t = \frac{12 \pm \sqrt{172}}{14}$ Since $\sqrt{172} = \sqrt{4 \times 43} = 2\sqrt{43}$: $t = \frac{12 \pm 2\sqrt{43}}{14}$ $t = \frac{6 \pm \sqrt{43}}{7}$ So, we have two possible values for $t = \tan(2x)$:

1. $\tan(2x) = \frac{6 + \sqrt{43}}{7}$
2. $\tan(2x) = \frac{6 - \sqrt{43}}{7}$

For the first case, $\tan(2x) = \frac{6 + \sqrt{43}}{7}$: The general solution is $2x = n\pi + \arctan\left(\frac{6 + \sqrt{43}}{7}\right)$, where $n$ is an integer. Dividing by 2, we get: $x = \frac{n\pi}{2} + \frac{1}{2} \arctan\left(\frac{6 + \sqrt{43}}{7}\right)$

For the second case, $\tan(2x) = \frac{6 - \sqrt{43}}{7}$: The general solution is $2x = m\pi + \arctan\left(\frac{6 - \sqrt{43}}{7}\right)$, where $m$ is an integer. Dividing by 2, we get: $x = \frac{m\pi}{2} + \frac{1}{2} \arctan\left(\frac{6 - \sqrt{43}}{7}\right)$

These two sets of solutions represent all real values of $x$ satisfying the given equation. The final answer is $x = \frac{n\pi}{2} + \frac{1}{2} \arctan\left(\frac{6 \pm \sqrt{43}}{7}\right), n \in \mathbb{Z}$.