Find-the-value-of-x-4-cos-x-3-sin-x-2-

Question Number 58804 by Tawa1 last updated on 30/Apr/19

Find the value of x : 4 \cos x + 3 \sin x = 2

Commented by MJS last updated on 30/Apr/19

generally

a \cos x + b \sin x = c

x = 2 \arctan t

t^{2} - \frac{2 b}{a + c} t - \frac{a - c}{a + c} = 0

t = \frac{b \pm \sqrt{a^{2} + b^{2} - c^{2}}}{a + c}

x = 2 π n + 2 \arctan \frac{b \pm \sqrt{a^{2} + b^{2} - c^{2}}}{a + c}; n \in Z

even if t \notin R

Answered by MJS last updated on 30/Apr/19

4 \cos x + 3 \sin x = 2

x = 2 \arctan t

4 \frac{1 - t^{2}}{1 + t^{2}} + 3 \frac{2 t}{1 + t^{2}} = 2

t^{2} - t - \frac{1}{3} = 0

t = \frac{1}{2} \pm \frac{\sqrt{21}}{6}

x = 2 π n + 2 \arctan \frac{3 \pm \sqrt{21}}{6}; n \in Z

Commented by Tawa1 last updated on 30/Apr/19

God bless you sir

Commented by Tawa1 last updated on 30/Apr/19

Sir, why when n = 1, the equation is not equal to 2

Answered by mr W last updated on 30/Apr/19

4 \cos x + 3 \sin x = 2

\frac{4}{5} \cos x + \frac{3}{5} \sin x = \frac{2}{5}

\cos α \cos x + \sin α \sin x = \frac{2}{5}

\cos (x - α) = \frac{2}{5}

\Rightarrow x - α = 2 n π \pm \cos^{- 1} \frac{2}{5}

\Rightarrow x = 2 n π + α \pm \cos^{- 1} \frac{2}{5}

\Rightarrow x = 2 n π + \cos^{- 1} \frac{4}{5} \pm \cos^{- 1} \frac{2}{5}

Commented by Tawa1 last updated on 30/Apr/19

God bless you sir

Commented by Tawa1 last updated on 30/Apr/19

Sir, when n = 1, the equation is not equal to 2 .

Commented by mr W last updated on 30/Apr/19

h o w d i d y o u c o m e t o t h a t ?

w i t h n = 1 :

x = 2 π + \cos^{- 1} \frac{4}{5} + \cos^{- 1} \frac{2}{5}

o r

x = 2 π + \cos^{- 1} \frac{4}{5} - \cos^{- 1} \frac{2}{5}

w i t h x = 2 π + \cos^{- 1} \frac{4}{5} + \cos^{- 1} \frac{2}{5} :

\cos x = \cos (\cos^{- 1} \frac{4}{5} + \cos^{- 1} \frac{2}{5})

= \frac{4}{5} \times \frac{2}{5} - \frac{3}{5} \times \frac{\sqrt{21}}{5} = \frac{8 - 3 \sqrt{21}}{25}

\sin x = \sin (\cos^{- 1} \frac{4}{5} + \cos^{- 1} \frac{2}{5})

= \frac{3}{5} \times \frac{2}{5} + \frac{4}{5} \times \frac{\sqrt{21}}{5} = \frac{6 + 4 \sqrt{21}}{25}

4 \cos x + 3 \sin x = \frac{4 (8 - 3 \sqrt{21}) + 3 (6 + 4 \sqrt{21})}{25}

= \frac{32 + 18}{25} = \frac{50}{25} = 2!

w i t h x = 2 π + \cos^{- 1} \frac{4}{5} - \cos^{- 1} \frac{2}{5} :

\cos x = \cos (\cos^{- 1} \frac{4}{5} - \cos^{- 1} \frac{2}{5})

= \frac{4}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{\sqrt{21}}{5} = \frac{8 + 3 \sqrt{21}}{25}

\sin x = \sin (\cos^{- 1} \frac{4}{5} - \cos^{- 1} \frac{2}{5})

= \frac{3}{5} \times \frac{2}{5} - \frac{4}{5} \times \frac{\sqrt{21}}{5} = \frac{6 - 4 \sqrt{21}}{25}

4 \cos x + 3 \sin x = \frac{4 (8 + 3 \sqrt{21}) + 3 (6 - 4 \sqrt{21})}{25}

= \frac{32 + 18}{25} = \frac{50}{25} = 2!

Commented by Tawa1 last updated on 30/Apr/19

Sir, when i used π = 360, i got 2 now . Before i used π = 3.142

Commented by MJS last updated on 30/Apr/19

maybe on your calculator you must set the

mode for the angle

usually {it}^{'} s “ {d e g}^{″} for degree (full circle = 360 °)

and “ {r a d}^{″} for radiant (full circle = 2 π)

Commented by Tawa1 last updated on 30/Apr/19

Ohh . Thanks for your time sir

Commented by MJS last updated on 30/Apr/19

{you}^{'} re welcome

Facebook Tweet Pin