cosx-sin3x-find-x-with-solution-pllllllz-

Question Number 60536 by hovea cw last updated on 21/May/19

cosx = \sin 3 x

find x with solution

pllllllz

Answered by mr W last updated on 21/May/19

\cos x = \sin (\frac{π}{2} - x) = \sin 3 x

\Rightarrow 3 x = \frac{π}{2} - x + 2 n π

\Rightarrow x = \frac{n π}{2} + \frac{π}{8}, n ϵ Z

o r

\Rightarrow 3 x = (2 n + 1) π - (\frac{π}{2} - x)

\Rightarrow x = n π + \frac{π}{4}, n ϵ Z

Commented by hovea cw last updated on 21/May/19

ok sir thanks

Answered by MJS last updated on 21/May/19

\cos x = \sin 3 x

x = \arctan t

\frac{1}{\sqrt{t^{2} + 1}} = \frac{t (3 - t^{2})}{\sqrt{{(t^{2} + 1)}^{3}}}

t^{2} + 1 = t (3 - t^{2})

t^{3} + t^{2} - 3 t + 1 = 0

(t - 1) (t^{2} + 2 t - 1) = 0

t_{1} = - 1 - \sqrt{2}; t_{2} = - 1 + \sqrt{2}; t_{3} = 1

x_{1} = \arctan (- 1 - \sqrt{2}) = - \frac{3 π}{8} \pm π n

x_{2} = \arctan (- 1 + \sqrt{2}) = \frac{π}{8} \pm π n

x_{3} = \arctan 1 = \frac{π}{4} \pm π n

[n \in N]

Commented by hovea cw last updated on 21/May/19

how did u get arctant = x

skr

Commented by MJS last updated on 21/May/19

I substituted it to get a solveable equation

in t

Commented by MJS last updated on 22/May/19

\sin 3 x = - \sin x (1 - {4 \cos}^{2} x)

\sin \arctan t = \frac{t}{\sqrt{t^{2} + 1}}

\cos \arctan t = \frac{1}{\sqrt{t^{2} + 1}}

\sin 3 x = - \frac{t}{\sqrt{t^{2} + 1}} (1 - \frac{4}{t^{2} + 1}) = \frac{t (3 - t^{2})}{\sqrt{{(t^{2} + 1)}^{3}}}

Answered by malwaan last updated on 22/May/19

c o s x = c o s (\frac{π}{2} - 3 x)

x = \pm (\frac{π}{2} - 3 x) + 2 n π

∴ x = \frac{π}{8} + \frac{n π}{2}; n \in Z

o r x = \frac{π}{4} - n π = \frac{π}{4} + n π; n \in Z

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