Question

25^2cos(x+3pi/2)=(1/5)^2cos(x+pi)

Answer 1

$25^{2cos(\frac{3\pi}{2}+x)}=(\frac{1}{5})^{2cos(\pi+x)}\\(5^2)^{2cos(\frac{3\pi}{2}+x)}=(5^2)^{-cos(\pi+x)}\\2cos(\frac{3\pi}{2}+x)+cos(\pi+x)=0\\2sinx-cosx=0\\2sinx=cosx|:cosx\\2tgx=1\\tgx=\frac{1}{2}\to x=arctg\frac{1}{2}+\pi n,n\in Z$

Answer 2

To solve the equation $25^2 \cos(x+\frac{3\pi}{2}) = \left(\frac{1}{5}\right)^2 \cos(x+\pi)$, you can follow these steps:

Step 1: Simplify both sides of the equation.

• $25^2 \cos(x+\frac{3\pi}{2}) = 625 \cos(x+\frac{3\pi}{2})$
• $\left(\frac{1}{5}\right)^2 \cos(x+\pi) = \frac{1}{25} \cos(x+\pi)$

So the equation becomes: $625 \cos(x+\frac{3\pi}{2}) = \frac{1}{25} \cos(x+\pi)$

Step 2: Use the properties of cosine function to simplify further.

• $\cos(x+\frac{3\pi}{2}) = -\sin(x)$
• $\cos(x+\pi) = -\cos(x)$

Now, the equation becomes: $625 (-\sin(x)) = \frac{1}{25} (-\cos(x))$

Step 3: Solve for $x$. Multiply both sides of the equation by $-25$ to eliminate the fractions: $625 \cdot 25 \sin(x) = \cos(x)$

$15625 \sin(x) = \cos(x)$

Now, you can use the identity $\sin^2(x) + \cos^2(x) = 1$ to solve for $x$: $\sin^2(x) + \cos^2(x) = 1$ $15625 \sin^2(x) + 15625 \cos^2(x) = 15625$ $15625 (\sin^2(x) + \cos^2(x)) = 15625$ $15625 = 15625$

Since this is a true statement, it means that the original equation is satisfied for all values of $x$. In other words, the equation is an identity and holds true for any value of $x$.