Question

2cos^ 5x-√2sin (п/2+5x)=0

Answer 1

Sin( $\frac{ \pi }{2}$+5x=cos5x
tak znachit
2 $cos^{2} 5x$- $\sqrt{2}$cos5x=0

2cos5x(cos5x- $\frac{ \sqrt{2} }{2}$)=0
cos5x=0
cos5x= $\frac{ \sqrt{2} }{2}$ 
5x= $\frac{ \pi }{2} + \pi n$ , n €Z 
x= $\frac{ \pi }{10} + \frac{\pi n}{10}$ , n €Z 
5x= $\frac{ \pi }{4} + \pi n$ , n €Z 
x= $\frac{ \pi }{40} + \frac{\pi n}{40}$ , n €Z 

Answer 2

To solve the trigonometric equation $2\cos^5(x) - \sqrt{2}\sin\left(\frac{\pi}{2} + 5x\right) = 0$, we can use some trigonometric identities to simplify and solve it step by step.

First, let's rewrite $\sin\left(\frac{\pi}{2} + 5x\right)$ using the angle sum identity for sine:

$\sin\left(\frac{\pi}{2} + 5x\right) = \sin\frac{\pi}{2}\cos(5x) + \cos\frac{\pi}{2}\sin(5x) = \cos(5x).$

Now, we have the equation:

$2\cos^5(x) - \sqrt{2}\cos(5x) = 0.$

Next, we can factor out $\cos(x)$ from both terms:

$\cos(x)(2\cos^4(x) - \sqrt{2}\cos(4x)) = 0.$

Now, we have two possible cases to consider:

Case 1: $\cos(x) = 0$

This implies $x = \frac{\pi}{2} + n\pi$ for some integer $n$, since $\cos(\frac{\pi}{2} + n\pi) = 0$ for all integers $n$.

Case 2: $2\cos^4(x) - \sqrt{2}\cos(4x) = 0$

Let's solve this equation separately:

$2\cos^4(x) - \sqrt{2}\cos(4x) = 0.$

Divide both sides by 2:

$\cos^4(x) - \frac{\sqrt{2}}{2}\cos(4x) = 0.$

Now, let $u = \cos(x)$, so we have:

$u^4 - \frac{\sqrt{2}}{2}\cos(4x) = 0.$

Now, we need to find the solutions for $u$. Let's focus on the second term:

$\cos(4x)$ can be expressed in terms of $u$ using a double angle identity:

$\cos(4x) = 2\cos^2(2x) - 1 = 2(2u^2 - 1)^2 - 1 = 8u^4 - 8u^2 + 1.$

So, our equation becomes:

$u^4 - \frac{\sqrt{2}}{2}(8u^4 - 8u^2 + 1) = 0.$

Now, let's solve for $u$:

$u^4 - 4\sqrt{2}u^4 + 4\sqrt{2}u^2 - \frac{\sqrt{2}}{2} = 0.$

Combine like terms:

$-3\sqrt{2}u^4 + 4\sqrt{2}u^2 - \frac{\sqrt{2}}{2} = 0.$