Solve $1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0$
Answer
365.7k+ 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.
Last updated date: 04th Oct 2023
•
Total views: 365.7k
•
Views today: 8.65k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

An alternating current can be produced by A a transformer class 12 physics CBSE

What is the value of 01+23+45+67++1617+1819+20 class 11 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers
