How do you solve $8 (sin x) (cos x) ({cos}^{2} x - {sin}^{2} x) = 1$ $8(sin x)(cos x)(cos^2 x - sin^2 x) = 1$ ?

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Answer 1 · 2016-06-18T15:31:30

$x = \frac{n π}{4} + {(- 1)}^{n} \cdot \frac{π}{24}$ $x=(npi)/4+(-1)^n*pi/24$

$8 (sin x) (cos x) ({cos}^{2} x - {sin}^{2} x) = 1$ $8(sin x)(cos x)(cos^2 x - sin^2 x) = 1$

Applying identities $2 sin x cos x = sin 2 x and ({cos}^{2} x - {sin}^{2} x) = cos 2 x$ $2sinx cosx =sin2x and (cos^2 x - sin^2 x)=cos2x$ we get

$4 sin 2 x cos 2 x = 1$ $4sin2xcos2x=1$

$\Rightarrow 2 sin 4 x = 1$ $=>2sin4x=1$

$\Rightarrow sin 4 x = \frac{1}{2} = sin (\frac{π}{6})$ $=>sin4x=1/2= sin(pi/6)$

$\Rightarrow 4 x = n π + {(- 1)}^{n} \cdot \frac{π}{6}$ $=>4x=npi+(-1)^n*pi/6$

$\Rightarrow x = \frac{n π}{4} + {(- 1)}^{n} \cdot \frac{π}{24}$ $=>x=(npi)/4+(-1)^n*pi/24$

How do you solve 8(sin x)(cos x)(cos^2 x - sin^2 x) = 18(sinx)(cosx)(cos2x−sin2x)=18(sin x)(cos x)(cos^2 x - sin^2 x) = 1?