solve-3-1-sin-x-3-1-cos-x-4-2-where-x-0-pi-2-

Question Number 119266 by benjo_mathlover last updated on 23/Oct/20

s o l v e \frac{\sqrt{3} - 1}{\sin x} + \frac{\sqrt{3} + 1}{\cos x} = 4 \sqrt{2}

w h e r e x \in (0, \frac{π}{2})

Commented by MJS_new last updated on 23/Oct/20

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

[sorry no time to type this transformation]

2 \sqrt{2} \sin (x + \frac{π}{12}) = 2 \sqrt{2} \sin 2 x

\sin (x + \frac{π}{12}) = \sin 2 x

obviously one solution is

x + \frac{π}{12} = 2 x \Rightarrow x = \frac{π}{12}

generally

\sin 2 x = \sin (x + \frac{π}{12})

x = \frac{π}{12} + 2 n π \lor x = \frac{11 π}{36} + \frac{2 n}{3} π

\Rightarrow for 0 ⩽ x < \frac{π}{2} we get

x = \frac{π}{12} \lor x = \frac{11 π}{36}

Commented by benjo_mathlover last updated on 23/Oct/20

t h a n k y o u a l l

Answered by Bird last updated on 23/Oct/20

\Rightarrow (\sqrt{3} - 1) c o s x + (\sqrt{3} + 1) s i n x

= 4 \sqrt{2} s i n x c o s x w e h a v e

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

\sqrt{4 - 2 \sqrt{3} + 4 + 2 \sqrt{3}} = 2 \sqrt{2}

e \Rightarrow 2 \sqrt{2 {} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) c o s x + \frac{\sqrt{3} + 1}{2 \sqrt{2}} s i n x}

= 2 \sqrt{2} s i n (2 x) \exists θ / s i n θ = \frac{\sqrt{3} - 1}{2 \sqrt{2}}

s n d c o s θ = \frac{\sqrt{3} + 1}{2 \sqrt{2}} \Rightarrow θ = a r c t a n (\frac{\sqrt{3} - 1}{\sqrt{3} + 1})

\Rightarrow s i n θ c o s x + c o s θ s i n x = s i n (2 x)

\Rightarrow s i n (θ + x) = s i n (2 x) \Rightarrow

θ + x = 2 x + 2 k π o r θ + x = π - 2 x + 2 k π \Rightarrow

x = θ - 2 k π o r 3 x = π - θ + 2 k π \Rightarrow

x = θ - 2 k π o r x = \frac{π - θ}{3} + \frac{2 k π}{3}

Answered by bemath last updated on 23/Oct/20

\frac{\sqrt{3} - 1}{4 \sqrt{2}} \cos x + \frac{\sqrt{3} + 1}{4 \sqrt{2}} \sin x = \frac{1}{2} \sin 2 x

\frac{\sqrt{3} - 1}{2 \sqrt{2}} \cos x + \frac{\sqrt{3} + 1}{2 \sqrt{2}} \sin x = \sin 2 x

\frac{\sqrt{6} - \sqrt{2}}{4} \cos x + \frac{\sqrt{6} + \sqrt{2}}{4} \sin x = \sin 2 x

\sin 15 ° \cos x + \cos 15 ° \sin x = \sin 2 x

\sin (x + 15 °) = \sin 2 x

\to {\begin{cases} 2 x = x + 15 ° + 2 n π \\ 2 x = 180 ° - x - 15 ° + 2 n π \end{cases}

\to {\begin{cases} x = 15 ° \\ x = 55 ° \end{cases}

Commented by MJS_new last updated on 23/Oct/20

great!

Commented by bemath last updated on 23/Oct/20

t h a n k y o u s i r

Answered by 1549442205PVT last updated on 23/Oct/20

\frac{\sqrt{3} - 1}{\sin x} + \frac{\sqrt{3} + 1}{\cos x} = 4 \sqrt{2} . Since two sides

are positive, squaring both two sides

we get the equivalent equation

\frac{4 - 2 \sqrt{3}}{\sin^{2} x} + \frac{4 + 2 \sqrt{3}}{\cos^{2} x} + \frac{4}{sinxcosx} = 32

\Leftrightarrow (2 - \sqrt{3}) (1 + \frac{1}{\tan^{2} x}) + (2 + \sqrt{3}) (1 + \tan^{2}) + \frac{2 (1 + \tan^{2} x)}{tanx} = 16

\Leftrightarrow \frac{(2 - \sqrt{3}) (1 + t^{2})}{t^{2}} + (2 + \sqrt{3}) (1 + t^{2}) + \frac{2 (1 + t^{2})}{t} = 16 (with t = tanx)

\Leftrightarrow (2 + \sqrt{3}) t^{4} + {2 t}^{3} - {12 t}^{2} + 2 t + 2 - \sqrt{3} = 0

\Leftrightarrow [t - (2 - \sqrt{3})] [(2 + \sqrt{3}) t^{3} + {3 t}^{2} - 3 (2 + \sqrt{3}) t - 1] = 0

i) t - (2 - \sqrt{3}) = 0 \Rightarrow tanx = t = 2 - \sqrt{3}

\Rightarrow x = 15 °

ii) (2 + \sqrt{3}) t^{3} + {3 t}^{2} - 3 (2 + \sqrt{3}) t - 1 = 0

\Leftrightarrow t = 1.428148, two negative roots rejected

\Rightarrow tanx = 1.428148 \Rightarrow x = 55 °

Thus, the given equation has two roots

x = 15 ° and x = 55 °