Question

1+cosx=sin^2x need help

Answer 1

Ответ:

Пошаговое объяснение:

1+cosx=sin²x

1+cosx=1-cos²x

cos²x+cosx=0

cos x*(cos x+1)=0

cos x=0    x= π/2 +πk

cos x+1=0   cos x= -1   x=π+2πk

Answer 2

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

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

2. Rewrite $\sin^2(x)$ using the Pythagorean identity: $\sin^2(x) = 1 - \cos^2(x)$.

   Now, your equation becomes: $1 + \cos(x) = 1 - \cos^2(x)$

3. Move all terms to one side of the equation to set it equal to zero: $\cos^2(x) + \cos(x) - 1 = 0$

4. This is a quadratic equation in terms of $\cos(x)$. You can solve it using the quadratic formula: $\cos(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   In this case, $a = 1$, $b = 1$, and $c = -1$, so plug these values into the quadratic formula:

   $\cos(x) = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$

   Simplify it further:

   $\cos(x) = \frac{-1 \pm \sqrt{5}}{2}$

5. Now, you have two possible values for $\cos(x)$:

   • $\cos(x) = \frac{-1 + \sqrt{5}}{2}$
   • $\cos(x) = \frac{-1 - \sqrt{5}}{2}$

6. To find the values of $x$, take the inverse cosine (arccosine) of both solutions:

   • $x = \arccos\left(\frac{-1 + \sqrt{5}}{2}\right)$
   • $x = \arccos\left(\frac{-1 - \sqrt{5}}{2}\right)$

7. Use a calculator to approximate these values. Keep in mind that the range of the arccosine function is between 0 and $\pi$ (0 and 180 degrees).

So, the solutions for $x$ are the values you get after evaluating the arccosine expressions for $\frac{-1 + \sqrt{5}}{2}$ and $\frac{-1 - \sqrt{5}}{2}$.