Question

1+cos(x^2+1)=sin^2(x^2+1)

Answer 1

$1+cos(x^2+1)=sin^2(x^2+1)\\x^2+1=t\\1+cos(t)=sin^2(t)\\1+cos(t)=1-cos^2(t)\\cos^2(t)+cos(t)=0\\cos(t)(cos(t)+1)=0\\cos(t)=0\\t= \frac{\pi}{2}+\pi k\\x^2+1=\frac{\pi}{2}+\pi k\\x^2=\frac{\pi-2}{2} +\pi k\\x=+- \sqrt{\frac{\pi-2}{2}+\pi k } \\cos(t)=-1\\t=\pi +2\pik\\x^2+1=\pi +2\pi k\\x^2=\pi-2\pi k\\x=+- \sqrt{\pi-1+2\pi k}$

Answer 2

To solve the equation 1 + cos(x^2 + 1) = sin^2(x^2 + 1), you can follow these steps:

1. Rewrite sin^2(x^2 + 1) in terms of cos(x^2 + 1) using the trigonometric identity: sin^2(x) = 1 - cos^2(x).

2. Substitute this identity into the equation:

1 + cos(x^2 + 1) = 1 - cos^2(x^2 + 1)

3. Now, you have a quadratic equation in terms of cos(x^2 + 1). Move all terms to one side of the equation:

cos^2(x^2 + 1) + cos(x^2 + 1) - 1 = 0

4. This is a quadratic equation in cos(x^2 + 1). You can solve it using the quadratic formula:

cos(x^2 + 1) = [-b ± √(b^2 - 4ac)] / (2a)

In this case, a = 1, b = 1, and c = -1. Plug these values into the formula:

cos(x^2 + 1) = [-1 ± √(1 + 4)] / 2

cos(x^2 + 1) = [-1 ± √5] / 2

5. Now, you have two possible values for cos(x^2 + 1):

a) cos(x^2 + 1) = (-1 + √5) / 2 b) cos(x^2 + 1) = (-1 - √5) / 2

6. To find the solutions for x, you'll need to take the inverse cosine (arccos) of these values:

a) x^2 + 1 = arccos((-1 + √5) / 2) b) x^2 + 1 = arccos((-1 - √5) / 2)

7. Solve for x in each equation:

a) x^2 = arccos((-1 + √5) / 2) - 1 b) x^2 = arccos((-1 - √5) / 2) - 1

8. To find the possible values of x, take the square root of both sides, but remember to consider both the positive and negative square roots:

a) x = ± √[arccos((-1 + √5) / 2) - 1] b) x = ± √[arccos((-1 - √5) / 2) - 1]

These are the solutions for the equation 1 + cos(x^2 + 1) = sin^2(x^2 + 1). However, the exact numerical values of these solutions would require a calculator or a numerical approximation method.

Answer 3

To solve the equation $1 + \cos(x^2 + 1) = \sin^2(x^2 + 1)$, you can use some trigonometric identities and algebraic manipulations. Here's how you can solve it step by step:

1. Start with the given equation: $1 + \cos(x^2 + 1) = \sin^2(x^2 + 1)$

2. Use the trigonometric identity $\sin^2(\theta) = 1 - \cos^2(\theta)$: $1 + \cos(x^2 + 1) = 1 - \cos^2(x^2 + 1)$

3. Move the $\cos^2(x^2 + 1)$ term to the left side of the equation: $\cos^2(x^2 + 1) + \cos(x^2 + 1) - 1 = 0$

4. Now, you have a quadratic equation in terms of $\cos(x^2 + 1)$. Let's make a substitution to simplify it. Let $y = \cos(x^2 + 1)$, so the equation becomes: $y^2 + y - 1 = 0$

5. Solve the quadratic equation for $y$. You can use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ In this case, $a = 1$, $b = 1$, and $c = -1$: $y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$ $y = \frac{-1 \pm \sqrt{5}}{2}$

6. Now that you've found the possible values of $y$, remember that $y = \cos(x^2 + 1)$. So you have: $\cos(x^2 + 1) = \frac{-1 \pm \sqrt{5}}{2}$

7. Solve for $x$: $x^2 + 1 = \cos^{-1}\left(\frac{-1 \pm \sqrt{5}}{2}\right)$

8. Since $x$ can take any real value, you may have multiple solutions for $x$. The solutions will depend on the specific values of $\frac{-1 \pm \sqrt{5}}{2}$.

So, you've now expressed the equation in terms of $x$. Depending on the values of $\frac{-1 \pm \sqrt{5}}{2}$, you'll have different solutions for $x$.