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Question: Solve \(1 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0\)...

Solve 1+sin⁡3x+cos⁡3x−32sin⁡2x=01 + {\sin ^3}x + {\cos ^3}x - \frac{3}{2}\sin 2x = 0

Explanation

Solution

Hint: Here, we will use the trigonometric formulas to simplify the given equation.

Given,
1+sin⁡3x+cos⁡3x−32sin⁡2x=0→(1)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\sin 2xi.e.., 2sin⁡xcos⁡x2\sin x\cos x.we get
⇒1+sin⁡3x+cos⁡3x−32sin⁡2x=0 ⇒1+sin⁡3x+cos⁡3x−32(2sin⁡xcos⁡x)=0 ⇒1+sin⁡3x+cos⁡3x−3sin⁡xcos⁡x=0 ⇒1+sin⁡3x+cos⁡3x−3(sin⁡x)(cos⁡x)(1)=0→(2) \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 a3+b3+c3−3abc=0{a^3} + {b^3} + {c^3} - 3abc = 0where a=1,b=sin⁡x,c=cos⁡xa = 1,b = \sin x,c = \cos x
And we now that
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca) ∴(a+b+c)(a2+b2+c2−ab−bc−ca)=0 \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=0a + b + c = 0 as the other factor is non-zero. Hence, from
equation (2), we can write
1+sin⁡x+cos⁡x=0→(3)1 + \sin x + \cos x = 0 \to (3)
Now, let us simplify equation (3) to find the values of ‘x’
⇒1+sin⁡x+cos⁡x=0 ⇒sin⁡x+cos⁡x=−1 \begin{gathered} \Rightarrow 1 + \sin x + \cos x = 0 \\\ \Rightarrow \sin x + \cos x = - 1 \\\ \end{gathered}
Let us multiply the above equation with 12\frac{1}{{\sqrt 2 }}we get,
⇒sin⁡x+cos⁡x=−1 ⇒(12)(sin⁡x+cos⁡x)=−12 ⇒12(sin⁡x)+12(cos⁡x)=−12 ⇒sin⁡xsin⁡(π4)+(cos⁡x)cos⁡(π4)=cos⁡(3π4)→(4)[∵sin⁡(π4)=12,cos⁡(π4)=12,cos⁡(3π4)=−12] \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⁡Asin⁡B+cos⁡Acos⁡B=cos⁡(A−B)\sin A\sin B + \cos A\cos B = \cos (A - B)where

A=xandB=π4A = x and B = \frac{\pi }{4}.Now let us apply the formulae ofsin⁡Asin⁡B+cos⁡Acos⁡B\sin A\sin B + \cos A\cos B we get

\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.