# How do you solve sin^2 = 2cosx - 2 algebraically?

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Nov 23, 2017

$x = \textcolor{b l u e}{2 n \pi}$ ; where $\textcolor{red}{n \in \mathbb{Z}}$ , the set of integers.

#### Explanation:

I am assuming the question is :

${\sin}^{2} x = 2 \cos x - 2$

We know the trigonometric identity$\rightarrow$

$\implies \textcolor{red}{{\sin}^{2} x + {\cos}^{2} x = 1}$

Subtract ${\cos}^{2} x$ from both sides to get :

$\implies {\sin}^{2} x = 1 - {\cos}^{2} x$

Put this value of ${\sin}^{2} x$ in the given equation :

$\implies 1 - {\cos}^{2} x = 2 \cos x - 2$

Add ${\cos}^{2} x$ both sides , we get :

$\implies 1 - \cancel{{\cos}^{2} x} + \cancel{{\cos}^{2} x} = {\cos}^{2} x + 2 \cos x - 2$

Now subtract $1$ from both sides :

$\implies \cancel{1} - \cancel{1} = {\cos}^{2} x + 2 \cos x - 2 - 1$

Now we have a quadratic equation in $\textcolor{b l u e}{{\cos}^{2} x} \rightarrow$

$\implies {\cos}^{2} x + 2 \cos x - 3 = 0$

We solve it by factorisation method (splitting the middle term) :

$\implies {\cos}^{2} x + 3 \cos x - \cos x - 3 = 0$

$\implies \cos x \left(\cos x + 3\right) - 1 \left(\cos x + 3\right) = 0$

$\implies \left(\cos x - 1\right) \left(\cos x + 3\right) = 0$

This gives us two solutions :

• $\cos x = - 3$ (rejected) , because color(blue)(cosx in [-1,+1] .
• $\cos x = 1$ (accepted)

The second one is the solution ,

$\therefore \textcolor{red}{\cos x = 1}$ , which has the general solution$\rightarrow$

$\implies \textcolor{b l u e}{x = 2 n \pi}$ , where $\textcolor{b l u e}{n \in \mathbb{Z}}$ .

Moreover you can solve it graphically also $\rightarrow$

Here are the curves of $\textcolor{red}{y = {\sin}^{2} x}$ and $\textcolor{b l u e}{y = 2 \cos x - 2} \rightarrow$

Note : The colors of the graphs have been mentioned.

Fun Fact : Here $\text{x-axis} \mathmr{and} y = 0$ is the common tangent to the
curves wherever they touch , i.e $\left(\forall x = 2 n \pi , n \in \mathbb{Z}\right)$ .

For further references , read this : solving trigonometric equations

Hope it helps : )

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