If $4\sin x.\cos y + 2\sin x + 2\cos y + 1 = 0$ where $x,y \in |0,2\pi |$ find the largest value of the sum $(x + y)$.
Answer
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Hint:
As we can see, they have given an equation and are asking us to find the largest sum. So, first try to reduce the given equation $4\sin x.\cos y + 2\sin x + 2\cos y + 1 = 0$ by taking the common term, then we get $\sin $ and $\cos $term. By equating $\sin $ and $\cos $term to zero that is to the right hand side value we get the value of $x$ and $y$. By adding these two values we get the required answer.
Complete Step by Step Solution:
They have given an equation $4\sin x.\cos y + 2\sin x + 2\cos y + 1 = 0$ and asking us to find the largest value of sum $(x + y)$.
So to find the largest sum $(x + y)$first we need to know the value of $x$ and $y$.
To find value of $x$ and $y$first try to reduce the given equation $4\sin x.\cos y + 2\sin x + 2\cos y + 1 = 0$
We can write $4\sin x.\cos y$as $2\sin x.2\cos y$. So the above equation will become,
$2\sin x.2\cos y + 2\sin x + 2\cos y + 1 = 0$
Now, we need to separate $\sin $ and $\cos $term, by taking common.
$ \Rightarrow 2\cos y(2\sin x + 1) + 1(2\sin x + 1) = 0$
$ \Rightarrow (2\cos y + 1)(2\sin x + 1) = 0$
Therefore, $(2\cos y + 1) = 0$ or $(2\sin x + 1) = 0$
$ \Rightarrow \cos y = - \dfrac{1}{2}$ $ \Rightarrow \sin x = - \dfrac{1}{2}$
$ \Rightarrow y = \pi + \dfrac{\pi }{3}$ $ \Rightarrow x = 2\pi - \dfrac{\pi }{6}$
$ \Rightarrow y = \dfrac{{4\pi }}{3}$ $ \Rightarrow x = \dfrac{{11\pi }}{6}$
Now we have the values of $x$ and $y$ that is $x = \dfrac{{11\pi }}{6}$ and $y = \dfrac{{4\pi }}{3}$.
By using the values of $x$ and $y$we need to calculate the largest value of sum $(x + y)$.
Therefore, by adding the values of $x$ and $y$as below, we get
$x + y = \dfrac{{11\pi }}{6} + \dfrac{{4\pi }}{3}$
Now we need to take the L.C.M. for the above equation to find the largest sum. We get,
$ \Rightarrow x + y = \dfrac{{11\pi + 8\pi }}{6}$
On simplifying the above equation, we get
$ \Rightarrow x + y = \dfrac{{19\pi }}{6}$
Therefore, the largest value of $x + y$ is $\dfrac{{19\pi }}{6}$.
Note:
Whenever they ask for the largest value of sum by giving an equation then it is as easy because just by simplifying the given equation we get the values of variables, then by adding these two variables we can arrive at the solution.
As we can see, they have given an equation and are asking us to find the largest sum. So, first try to reduce the given equation $4\sin x.\cos y + 2\sin x + 2\cos y + 1 = 0$ by taking the common term, then we get $\sin $ and $\cos $term. By equating $\sin $ and $\cos $term to zero that is to the right hand side value we get the value of $x$ and $y$. By adding these two values we get the required answer.
Complete Step by Step Solution:
They have given an equation $4\sin x.\cos y + 2\sin x + 2\cos y + 1 = 0$ and asking us to find the largest value of sum $(x + y)$.
So to find the largest sum $(x + y)$first we need to know the value of $x$ and $y$.
To find value of $x$ and $y$first try to reduce the given equation $4\sin x.\cos y + 2\sin x + 2\cos y + 1 = 0$
We can write $4\sin x.\cos y$as $2\sin x.2\cos y$. So the above equation will become,
$2\sin x.2\cos y + 2\sin x + 2\cos y + 1 = 0$
Now, we need to separate $\sin $ and $\cos $term, by taking common.
$ \Rightarrow 2\cos y(2\sin x + 1) + 1(2\sin x + 1) = 0$
$ \Rightarrow (2\cos y + 1)(2\sin x + 1) = 0$
Therefore, $(2\cos y + 1) = 0$ or $(2\sin x + 1) = 0$
$ \Rightarrow \cos y = - \dfrac{1}{2}$ $ \Rightarrow \sin x = - \dfrac{1}{2}$
$ \Rightarrow y = \pi + \dfrac{\pi }{3}$ $ \Rightarrow x = 2\pi - \dfrac{\pi }{6}$
$ \Rightarrow y = \dfrac{{4\pi }}{3}$ $ \Rightarrow x = \dfrac{{11\pi }}{6}$
Now we have the values of $x$ and $y$ that is $x = \dfrac{{11\pi }}{6}$ and $y = \dfrac{{4\pi }}{3}$.
By using the values of $x$ and $y$we need to calculate the largest value of sum $(x + y)$.
Therefore, by adding the values of $x$ and $y$as below, we get
$x + y = \dfrac{{11\pi }}{6} + \dfrac{{4\pi }}{3}$
Now we need to take the L.C.M. for the above equation to find the largest sum. We get,
$ \Rightarrow x + y = \dfrac{{11\pi + 8\pi }}{6}$
On simplifying the above equation, we get
$ \Rightarrow x + y = \dfrac{{19\pi }}{6}$
Therefore, the largest value of $x + y$ is $\dfrac{{19\pi }}{6}$.
Note:
Whenever they ask for the largest value of sum by giving an equation then it is as easy because just by simplifying the given equation we get the values of variables, then by adding these two variables we can arrive at the solution.
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