Question

Cos(6x) - sqrt(2)cos(5pi - 3x) + 1 = 0

Answer 1

cos6x+1=2cos²3x

По формулам приведения:

cos(5π-3x)=-сos3x

Уравнение:

2cos²3x+√2сos3x=0

cos3x·(2cos3x+√2)=0

cos3x=0  или  2 сos3x+√2=0

cos3x=0⇒3x=(π/2)+πk, k∈Z⇒x=(π/6)+(π/3)k, k∈Z

2 сos3x=-√2⇒cos3x=-√2/2⇒3x=(-3π/4)+2πn, n∈Z⇒x=(-π/4)+(2π/3)n, n∈Z

Answer 2

To solve the equation $\cos(6x) - \sqrt{2}\cos(5\pi - 3x) + 1 = 0$, we need to find the values of $x$ that satisfy the equation. Let's break it down step by step:

Step 1: Simplify $\cos(5\pi - 3x)$: Recall the cosine function's periodicity property: $\cos(\theta) = \cos(\theta + 2\pi n)$ for any integer $n$. In our case, we have $\cos(5\pi - 3x) = \cos(-3x + 5\pi)$. Using the periodicity property, we can rewrite this as follows: $\cos(5\pi - 3x) = \cos(-3x + 5\pi) = \cos(-3x + 5\pi + 2\pi) = \cos(-3x + 7\pi)$.

Step 2: Rewrite $\cos(6x)$ using the double angle formula for cosine: The double angle formula for cosine states that $\cos(2\theta) = 2\cos^2(\theta) - 1$. Applying this to $\cos(6x)$, we get: $\cos(6x) = 2\cos^2(3x) - 1$.

Step 3: Replace $\cos(6x)$ and $\cos(5\pi - 3x)$ in the equation: After the simplifications, our equation becomes: $2\cos^2(3x) - 1 - \sqrt{2}\cos(-3x + 7\pi) + 1 = 0$.

Step 4: Further simplify: Notice that $\cos(-\theta) = \cos(\theta)$, so $\cos(-3x + 7\pi) = \cos(3x - 7\pi)$. Our equation now becomes: $2\cos^2(3x) - \sqrt{2}\cos(3x - 7\pi) = 0$.

Step 5: Factor out $\cos(3x)$ from the equation: $2\cos^2(3x) - \sqrt{2}\cos(3x - 7\pi) = \cos(3x) \left(2\cos(3x) - \sqrt{2}\right) = 0$.

Step 6: Solve for $\cos(3x)$: To find the values of $x$, we can set each factor equal to zero:

1. $\cos(3x) = 0$ (This means $3x = \frac{\pi}{2} + 2\pi n$ for any integer $n$).
2. $2\cos(3x) - \sqrt{2} = 0$ (This means $2\cos(3x) = \sqrt{2}$ and $cos(3x) = \frac{\sqrt{2}}{2}$).

Now, let's solve each equation for $x$:

1. For $\cos(3x) = 0$: $3x = \frac{\pi}{2} + 2\pi n$ $x = \frac{\pi}{6} + \frac{2\pi n}{3}$ (where $n$ is an integer).

2. For $\cos(3x) = \frac{\sqrt{2}}{2}$: We know that $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$