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JEE MainMathematicsModerateTrigonometry

If 4sin⁡4x+cos⁡4x=1,x≠nπ 4\sin^{4}x + \cos^{4}x = 1,x \neq n\pi\ then x

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The given equation can be put in the form
4sin⁡4x=1−cos⁡4x4\sin^{4}x = 1 - \cos^{4}x
=(1−cos⁡2x)(1+cos⁡2x)= \left( 1 - \cos^{2}x \right)\left( 1 + \cos^{2}x \right)
 or
sin⁡2x[4sin⁡2x−1−(1−sin⁡2x)]=0\sin^{2}{x\left\lbrack 4\sin^{2}x - 1 - \left( 1 - \sin^{2}x \right) \right\rbrack} = 0
 or  sin⁡2x[5sin⁡2x−2]=0{\ \sin^{2}}{x\left\lbrack 5\sin^{2}x - 2 \right\rbrack} = 0
∴sinx=0 \therefore sinx = 0\ or  sinx=±2/5sinx = \pm \sqrt{2/5}

Source: Grass-G / M-18 Mathematics Paper 27 Nov Third.docx

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