Cos(p/2+x)-sin(p-x)=0

To solve the equation $\cos (\frac{\pi }{2}+x)-\sin (\pi -x)=0$ , we will use trigonometric reduction formulas (reduction identities). 1. Applying Reduction Formulas First, let's simplify each term using standard trigonometric identities:

• For the first term: $\cos (\frac{\pi }{2}+x)$
  Since $\frac{\pi }{2}$ is an "odd" multiple of $\frac{\pi }{2}$ , the function changes from cosine to sine. In the second quadrant, cosine is negative.
  $\cos (\frac{\pi }{2}+x)=-\sin \left(x\right)$ For the second term: $\sin (\pi -x)$
  Since $\pi$ is an "even" multiple of $\frac{\pi }{2}$ , the function remains sine. In the second quadrant, sine is positive.
  $\sin (\pi -x)=\sin \left(x\right)$

2. Substituting back into the Equation Now, substitute these simplified terms back into the original equation: $(-\sin (x\left)\right)-\left(\sin \right(x\left)\right)=0$Combine the like terms: $-2\sin \left(x\right)=0$3. Solving for x Divide both sides by $-2$: $\sin \left(x\right)=0$The general solution for the equation $\sin \left(x\right)=0$ is found at the points where the sine function crosses the horizontal axis on the unit circle ( $0,\pi ,2\pi ,\dots$): $x=\pi k,\text{where}k\in \mathbb{Z}$Final Answer: $x=\pi k,\text{where}k\text{is any integer.}$ Would you like me to check if specific solutions fall within a particular numerical interval?