# Solve $1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0$

Last updated date: 18th Mar 2023

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Answer

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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.

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