How do you solve $\sin 2x-\cos x-2\sin x+1=0$?
Answer
632.1k+ views
Hint: In the question we have the multiple angle which is $\sin 2x$. So, we will apply the known formula $\sin 2x=2\sin x\cos x$ in the given equation. Now we will take $\cos x$ as common from the terms $2\sin x\cos x$, $\cos x$ and simplify the obtained equation. Again, we will observe the terms in the obtained equation and simplify the equation and equate the terms in the equation to the zero. Now we will get two equations, from these equations we will solve each one by considering the trigonometric values of different ratios. Now we will get the required solution.
Complete step-by-step answer:
Given that $\sin 2x-\cos x-2\sin x+1=0$.
We have the trigonometric formula $\sin 2x=2\sin x\cos x$. Substituting this value in the given equation, then we will get
$\begin{align}
& \sin 2x-\cos x-2\sin x+1=0 \\
& \Rightarrow 2\sin x\cos x-\cos x-2\sin x+1=0 \\
\end{align}$
Taking $\cos x$ common from the terms $2\sin x\cos x$, $\cos x$ in the above equation, then we will get
$\Rightarrow \cos x\left( 2\sin x-1 \right)-2\sin x+1=0$
Taking $-1$ common from the terms $2\sin x$, $1$ in the above equation, then we will get
$\Rightarrow \cos x\left( 2\sin x-1 \right)-1\left( 2\sin x-1 \right)=0$
Again taking $2\sin x-1$ common from the above equation, then we will get
$\Rightarrow \left( 2\sin x-1 \right)\left( \cos x-1 \right)=0$
Equating each term individually to the zero, then we will have
$2\sin x-1=0$ or $\cos x-1=0$.
Simplifying the above equations then we will get
$\begin{align}
& 2\sin x-1=0 \\
& \Rightarrow 2\sin x=1 \\
& \Rightarrow \sin x=\dfrac{1}{2} \\
\end{align}$ or $\begin{align}
& \cos x-1=0 \\
& \Rightarrow \cos x=1 \\
\end{align}$
We have $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ and $\cos \pi =1$, then the solutions of above equations are
$x=n\pi +{{\left( - \right)}^{n}}\dfrac{\pi }{6}$ or $x=2k\pi $ where $n,k\in Z$.
Note: We can also plot a graph of the given equation with is given below
From this graph also we can find the solution of the given equation by observing the points where the given equation meets the $x-axis$. The points where the given equation meets the $x-axis$ are the solutions for the given equation.
Complete step-by-step answer:
Given that $\sin 2x-\cos x-2\sin x+1=0$.
We have the trigonometric formula $\sin 2x=2\sin x\cos x$. Substituting this value in the given equation, then we will get
$\begin{align}
& \sin 2x-\cos x-2\sin x+1=0 \\
& \Rightarrow 2\sin x\cos x-\cos x-2\sin x+1=0 \\
\end{align}$
Taking $\cos x$ common from the terms $2\sin x\cos x$, $\cos x$ in the above equation, then we will get
$\Rightarrow \cos x\left( 2\sin x-1 \right)-2\sin x+1=0$
Taking $-1$ common from the terms $2\sin x$, $1$ in the above equation, then we will get
$\Rightarrow \cos x\left( 2\sin x-1 \right)-1\left( 2\sin x-1 \right)=0$
Again taking $2\sin x-1$ common from the above equation, then we will get
$\Rightarrow \left( 2\sin x-1 \right)\left( \cos x-1 \right)=0$
Equating each term individually to the zero, then we will have
$2\sin x-1=0$ or $\cos x-1=0$.
Simplifying the above equations then we will get
$\begin{align}
& 2\sin x-1=0 \\
& \Rightarrow 2\sin x=1 \\
& \Rightarrow \sin x=\dfrac{1}{2} \\
\end{align}$ or $\begin{align}
& \cos x-1=0 \\
& \Rightarrow \cos x=1 \\
\end{align}$
We have $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ and $\cos \pi =1$, then the solutions of above equations are
$x=n\pi +{{\left( - \right)}^{n}}\dfrac{\pi }{6}$ or $x=2k\pi $ where $n,k\in Z$.
Note: We can also plot a graph of the given equation with is given below
From this graph also we can find the solution of the given equation by observing the points where the given equation meets the $x-axis$. The points where the given equation meets the $x-axis$ are the solutions for the given equation.
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