Questions & Answers
Question
AnswersSolve $1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0$
Answer
Verified150k+ views
Hint: Here, we will use the trigonometric formulas to simplify the given equation.
Given,
$1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0 \to (1)$
Now, let us simplify the equation (1) by substituting the formula of $\sin 2x$i.e.., $2\sin x\cos
x$.we get
$\begin{gathered}
\Rightarrow 1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0 \\
\Rightarrow 1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}(2\sin x\cos x) = 0 \\
\Rightarrow 1 + {\sin ^3}x + {\cos ^3}x - 3\sin x\cos x = 0 \\
\Rightarrow 1 + {\sin ^3}x + {\cos ^3}x - 3(\sin x)(\cos x)(1) = 0 \to (2) \\
\end{gathered} $
As, we can see equation (2) is in the form of ${a^3} + {b^3} + {c^3} - 3abc = 0$where $a = 1,b
= \sin x,c = \cos x$
And we now that
$\begin{gathered}
{a^3} + {b^3} + {c^3} - 3abc = (a + b + c)({a^2} + {b^2} + {c^2} - ab - bc - ca) \\
\therefore (a + b + c)({a^2} + {b^2} + {c^2} - ab - bc - ca) = 0 \\
\end{gathered} $
Here, we will consider the factor $a + b + c = 0$ as the other factor is non-zero. Hence, from
equation (2), we can write
$1 + \sin x + \cos x = 0 \to (3)$
Now, let us simplify equation (3) to find the values of ‘x’
$\begin{gathered}
\Rightarrow 1 + \sin x + \cos x = 0 \\
\Rightarrow \sin x + \cos x = - 1 \\
\end{gathered} $
Let us multiply the above equation with $\frac{1}{{\sqrt 2 }}$we get,
$\begin{gathered}
\Rightarrow \sin x + \cos x = - 1 \\
\Rightarrow (\frac{1}{{\sqrt 2 }})(\sin x + \cos x) = - \frac{1}{{\sqrt 2 }} \\
\Rightarrow \frac{1}{{\sqrt 2 }}(\sin x) + \frac{1}{{\sqrt 2 }}(\cos x) = - \frac{1}{{\sqrt 2 }} \\
\Rightarrow \sin x\sin (\frac{\pi }{4}) + (\cos x)\cos (\frac{\pi }{4}) = \cos (\frac{{3\pi }}{4}) \to (4)[\because \sin (\frac{\pi }{4}) = \frac{1}{{\sqrt 2 }},\cos (\frac{\pi }{4}) = \frac{1}{{\sqrt 2 }},\cos (\frac{{3\pi }}{4}) = - \frac{1}{{\sqrt 2 }}] \\
\end{gathered} $
As, we can see equation (4) is in the form of $\sin A\sin B + \cos A\cos B = \cos (A - B)$where
$A = x and B = \frac{\pi }{4}$.Now let us apply the formulae of$\sin A\sin B + \cos A\cos B$ we get
\[\begin{gathered}
\Rightarrow \cos (x - \frac{\pi }{4}) = \cos (\frac{{3\pi }}{4}) \\
\Rightarrow x - \frac{\pi }{4} = 2n\pi \pm \frac{{3\pi }}{4} \to (5),['n'{\text{is integral number]}} \\
\end{gathered} \]
Therefore, solving equation (5) we get,
$ \Rightarrow x = 2n\pi + \pi and x = 2n\pi - \frac{\pi }{2}['n'{\text{is integral number}}]$
Hence, the values of ‘x’ satisfying $1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0$is$x =
2n\pi + \pi and x = 2n\pi - \frac{\pi }{2}$.
Note: Here, we have added $'2n\pi '$to the $\frac{{3\pi }}{4}$after cancelling the cosine terms on the both sides as $'2\pi '$is the period of the cosine function and n is an integral
number.
Given,
$1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0 \to (1)$
Now, let us simplify the equation (1) by substituting the formula of $\sin 2x$i.e.., $2\sin x\cos
x$.we get
$\begin{gathered}
\Rightarrow 1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0 \\
\Rightarrow 1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}(2\sin x\cos x) = 0 \\
\Rightarrow 1 + {\sin ^3}x + {\cos ^3}x - 3\sin x\cos x = 0 \\
\Rightarrow 1 + {\sin ^3}x + {\cos ^3}x - 3(\sin x)(\cos x)(1) = 0 \to (2) \\
\end{gathered} $
As, we can see equation (2) is in the form of ${a^3} + {b^3} + {c^3} - 3abc = 0$where $a = 1,b
= \sin x,c = \cos x$
And we now that
$\begin{gathered}
{a^3} + {b^3} + {c^3} - 3abc = (a + b + c)({a^2} + {b^2} + {c^2} - ab - bc - ca) \\
\therefore (a + b + c)({a^2} + {b^2} + {c^2} - ab - bc - ca) = 0 \\
\end{gathered} $
Here, we will consider the factor $a + b + c = 0$ as the other factor is non-zero. Hence, from
equation (2), we can write
$1 + \sin x + \cos x = 0 \to (3)$
Now, let us simplify equation (3) to find the values of ‘x’
$\begin{gathered}
\Rightarrow 1 + \sin x + \cos x = 0 \\
\Rightarrow \sin x + \cos x = - 1 \\
\end{gathered} $
Let us multiply the above equation with $\frac{1}{{\sqrt 2 }}$we get,
$\begin{gathered}
\Rightarrow \sin x + \cos x = - 1 \\
\Rightarrow (\frac{1}{{\sqrt 2 }})(\sin x + \cos x) = - \frac{1}{{\sqrt 2 }} \\
\Rightarrow \frac{1}{{\sqrt 2 }}(\sin x) + \frac{1}{{\sqrt 2 }}(\cos x) = - \frac{1}{{\sqrt 2 }} \\
\Rightarrow \sin x\sin (\frac{\pi }{4}) + (\cos x)\cos (\frac{\pi }{4}) = \cos (\frac{{3\pi }}{4}) \to (4)[\because \sin (\frac{\pi }{4}) = \frac{1}{{\sqrt 2 }},\cos (\frac{\pi }{4}) = \frac{1}{{\sqrt 2 }},\cos (\frac{{3\pi }}{4}) = - \frac{1}{{\sqrt 2 }}] \\
\end{gathered} $
As, we can see equation (4) is in the form of $\sin A\sin B + \cos A\cos B = \cos (A - B)$where
$A = x and B = \frac{\pi }{4}$.Now let us apply the formulae of$\sin A\sin B + \cos A\cos B$ we get
\[\begin{gathered}
\Rightarrow \cos (x - \frac{\pi }{4}) = \cos (\frac{{3\pi }}{4}) \\
\Rightarrow x - \frac{\pi }{4} = 2n\pi \pm \frac{{3\pi }}{4} \to (5),['n'{\text{is integral number]}} \\
\end{gathered} \]
Therefore, solving equation (5) we get,
$ \Rightarrow x = 2n\pi + \pi and x = 2n\pi - \frac{\pi }{2}['n'{\text{is integral number}}]$
Hence, the values of ‘x’ satisfying $1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0$is$x =
2n\pi + \pi and x = 2n\pi - \frac{\pi }{2}$.
Note: Here, we have added $'2n\pi '$to the $\frac{{3\pi }}{4}$after cancelling the cosine terms on the both sides as $'2\pi '$is the period of the cosine function and n is an integral
number.