Question

Find all solutions of the equation in the interval [0,2?)Write your answer in radians in terms of ?4cosx=-sin^{2}x+1

Answer 1

Answer:

$\frac{\pi}{2}$ and $\frac{3\pi}{2}$.

Step-by-step explanation:

We have been given an equation $4\text{cos}(x)=-\text{sin}^{2}(x)+1$. We are asked to find all the solutions of our given equation on interval $[0,2\pi]$ in radians in terms of pi.

Let us solve our given equation using property $\text{sin}^{2}(x)=1-\text{cos}^{2}(x)$.

$4\text{cos}(x)=-\text{sin}^{2}(x)+1$

$4\text{cos}(x)=-(1-\text{cos}^{2}(x))+1$

$4\text{cos}(x)=-1+\text{cos}^{2}(x)+1$

$4\text{cos}(x)=\text{cos}^{2}(x)$

Subtract $\text{cos}^{2}(x)$ from both sides:

$4\text{cos}(x)-\text{cos}^{2}(x)=0$

Upon factoring out our equation, we will get:

$\text{cos}(x)(4-\text{cos}(x))=0$

Using zero product property, we will get:

$\text{cos}(x)=0\text{ (or) }(4-\text{cos}(x))=0$

$\text{cos}(x)=0\text{ (or) }\text{cos}(x)=4$

We know that cosine function oscillates between -1 and 1, therefore, there is no solution for $\text{cos}(x)=4$.

We also know that $cos(x)=0$ at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. Therefore, the general solution for our given equation would be $\frac{\pi}{2}+2\pi n$ and $\frac{3\pi}{2}+2\pi n$.

Since we need our solution on interval $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, so we can only use $n=0$ for both solutions. If we will use $n=1$, then we will be adding 2 pi to each $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, which will be out of our boundaries.

Therefore, the solution for our equation is $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.