JEE Main 2024 — Mathematics Trigonometry

Given: $\frac{3\mathrm{cos}2x+{\mathrm{cos}}^{3}2x}{{\mathrm{cos}}^{6}x-{\mathrm{sin}}^{6}x}={x}^{3}-{x}^{2}+6$cos6x−sin6x3cos2x+cos32x​=x3−x2+6

$$\Rightarrow \frac{\mathrm{cos}2x(3+{\mathrm{cos}}^{2}2x)}{{({\mathrm{cos}}^{2}x)}^{3}-{({\mathrm{sin}}^{2}x)}^{3}}={x}^{3}-{x}^{2}+6$$⇒(cos2x)3−(sin2x)3cos2x(3+cos22x)​=x3−x2+6

$$\Rightarrow \frac{\mathrm{cos}2x(3+{\mathrm{cos}}^{2}2x)}{({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x)({\mathrm{cos}}^{4}x+{\mathrm{sin}}^{4}x+{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x)}={x}^{3}-{x}^{2}+6$$⇒(cos2x−sin2x)(cos4x+sin4x+sin2xcos2x)cos2x(3+cos22x)​=x3−x2+6

$$\Rightarrow \frac{(3+{\mathrm{cos}}^{2}2x)}{{({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x)}^{2}-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}={x}^{3}-{x}^{2}+6$$⇒(sin2x+cos2x)2−2sin2xcos2x+sin2xcos2x(3+cos22x)​=x3−x2+6

$$\Rightarrow \frac{(3+{\mathrm{cos}}^{2}2x)}{1-{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}={x}^{3}-{x}^{2}+6$$⇒1−sin2xcos2x(3+cos22x)​=x3−x2+6

$$\Rightarrow \frac{(3+{\mathrm{cos}}^{2}2x)}{1-\frac{{(2\mathrm{sin}x\mathrm{cos}x)}^{2}}{4}}={x}^{3}-{x}^{2}+6$$⇒1−4(2sinxcosx)2​(3+cos22x)​=x3−x2+6

$$\Rightarrow 4(\frac{3+{\mathrm{cos}}^{2}2x}{4-{\mathrm{sin}}^{2}2x})={x}^{3}-{x}^{2}+6$$⇒4(4−sin22x3+cos22x​)=x3−x2+6

$$\Rightarrow 4(\frac{3+1-{\mathrm{sin}}^{2}2x}{4-{\mathrm{sin}}^{2}2x})={x}^{3}-{x}^{2}+6$$⇒4(4−sin22x3+1−sin22x​)=x3−x2+6

$$\Rightarrow 4={x}^{3}-{x}^{2}+6$$⇒4=x3−x2+6

$$\Rightarrow {x}^{3}-{x}^{2}+2=0$$⇒x3−x2+2=0

$$\Rightarrow (x+1)({x}^{2}-2x+2)=0$$⇒(x+1)(x2−2x+2)=0

$$\Rightarrow (x+1)({(x-1)}^{2}+1)=0$$⇒(x+1)((x−1)2+1)=0

So, the sum of real roots is $-1$−1.