Question

Sin^2(п/2-x)+cos x =0

Answer 1

$\sin {}^{2} ( \frac{\pi}{2} - x ) + \cos(x) = 0$

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

$\cos {}^{2} (x) + \cos(x) = 0 \\ \cos(x) ( \cos(x) + 1) = 0 \\ \cos(x) = 0 \\ \cos(x) = - 1 \\ x = \frac{\pi}{2} + \pi n \\ x = \pi + 2\pi n$

Answer 2

To solve the equation $\sin^2\left(\frac{\pi}{2} - x\right) + \cos(x) = 0$, we'll use trigonometric identities and algebraic manipulation.

1. Start with the given equation: $\sin^2\left(\frac{\pi}{2} - x\right) + \cos(x) = 0$

2. Use the identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to rewrite $\sin^2\left(\frac{\pi}{2} - x\right)$ as $1 - \cos^2\left(\frac{\pi}{2} - x\right)$: $1 - \cos^2\left(\frac{\pi}{2} - x\right) + \cos(x) = 0$

3. Use the identity $\cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta)$ to further simplify: $1 - \sin^2(x) + \cos(x) = 0$

4. Now, use the Pythagorean identity $1 - \sin^2(x) = \cos^2(x)$: $\cos^2(x) + \cos(x) = 0$

5. Factor out the common term $\cos(x)$: $\cos(x)(\cos(x) + 1) = 0$

Now, we have two possible cases to consider:

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

Case 2: $\cos(x) + 1 = 0$

Let's solve each case separately:

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

To solve for $x$ in this case, we know that $\cos(x) = 0$ when $x$ is an odd multiple of $\frac{\pi}{2}$, i.e., $x = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

Case 2: $\cos(x) + 1 = 0$

To solve for $x$ in this case, subtract 1 from both sides: $\cos(x) = -1$

The cosine function is equal to -1 at odd multiples of $\pi$, so: $x = (2n+1)\pi$

So, the solutions to the equation $\sin^2\left(\frac{\pi}{2} - x\right) + \cos(x) = 0$ are: $x = \frac{\pi}{2} + n\pi$ for Case 1 and $x = (2n+1)\pi$ for Case 2, where $n$ is an integer.