Mathematics › Trigonometric Equations › Solving Trigonometric Equation
The set of solutions of the system of equations $$ \begin{aligned} x+y & =\frac{2 \pi}{3} \\ \text { and }…
The set of solutions of the system of equations
$$
\begin{aligned}
x+y & =\frac{2 \pi}{3} \\
\text { and } \quad \cos x+\cos y & =\frac{3}{2},
\end{aligned}
$$
where $x, y$ are real, is
$\left\{(x, y): \cos \left(\frac{x-y}{2}\right)=\frac{1}{2}\right\}$ $\left\{(x, y): \sin \left(\frac{x-y}{2}\right)=\frac{1}{2}\right\}$ $\left\{(x, y): \cos (x-y)=\frac{1}{2}\right\}$ Empty set Solution
Given system of equation is
$
x+y=\frac{2 \pi}{3}
$
and $\cos x+\cos y=\frac{3}{2}$, where $x, y$ are real.
$
\begin{aligned}
& \Rightarrow \quad 2 \cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)=\frac{3}{2} \\
& \Rightarrow \quad \cos \left(\frac{1}{2} \cdot \frac{2 \pi}{3}\right) \cdot \cos \left(\frac{x-y}{2}\right)=\frac{3}{4}
\end{aligned}
$
[from Eq. (i)]
$
\begin{aligned}
\Rightarrow & \cos \left(\frac{\pi}{3}\right) \cdot \cos \left(\frac{x-y}{2}\right) & =\frac{3}{4} \\
\Rightarrow & \frac{1}{2} \cos \left(\frac{x-y}{2}\right) & =\frac{3}{4} \\
\Rightarrow & \cos \left(\frac{x-y}{2}\right) & =\frac{3}{2}
\end{aligned}
$
Now, we have
$
\begin{aligned}
& \cos (x-y)=2 \cos ^2\left(\frac{x-y}{2}\right)-1 \\
& \left.=2 \times \frac{9}{4}-1=\frac{9}{2}-1=\frac{7}{2} \quad \text { [from Eq. (ii) }\right] \\
& \text { and } \cos (x-y)=1-2 \sin ^2\left(\frac{x-y}{2}\right) \\
& \Rightarrow \quad \sin \left(\frac{x-y}{2}\right) < 0 \\
&
\end{aligned}
$
So, system of equation have empty set of solution
Asked in: AP EAMCET 2013
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