Question

solve the equation 3cos2x-4sin2x=2 0 11 years ago

Answer 1

there is a direct formula for representing `acosx + bsinx` as a single sinusoidal function

Answer 2

\[3(\cos^2x-\sin^2x)+4(2sinxcosx)=2\]
\[8sinxcosx+3-6\sin^2x=2\]

Answer 3

say 2x = t

\[3\cos t - 4 \sin t = \langle 3, -4\rangle \bullet \langle \cos t, \sin t \rangle \]

Answer 4

but isnt that a double angle?

Answer 5

lets hope the substitution works :) if it doesn't work, we can try the other method

Answer 6

\[= |\langle 3, -4\rangle | ~ | \langle \cos t, \sin t \rangle | \cos (t - \phi) \]

\[= 5 \cos (t - \phi)\]

\(\phi = \tan^{-1}(\frac{-4}{3})\)

Answer 7

Cos-1=2/5=66.42182 Cosine is positive in the 4th quadrant
360-66.42182=293.578
293.578+tan-1(-4/3)=346.708

Answer 8

66.42182+53.130=119.55

Answer 9

although iam not sure if my working is right or not?@ganeshie8

Answer 10

@ganeshie8 Could you help in solving with the other method if you are free?

Answer 11

solve :

\[5\cos\left(2x -\tan^{-1}(\frac{-4}{3})\right) = 2\]

Answer 12

\[2x -\tan^{-1}(\frac{-4}{3}) = \cos^{-1}(\frac{2}{5})\]

Answer 13

\[2x -(-53.13^{\circ}) = 66.42^{\circ}\]

Answer 14

\[2x = 66.42^{\circ} - 53.13^{\circ}\]

Answer 15

\[2x = 13.29^{\circ}\]

Answer 16

since cos(x) = cos( \(\color{red}{-}\)x), the general solutions are :

\[2x = 13.29^{\circ} + 360 n\] and
\[2x = \color{red}{-}13.29^{\circ} + 360 n\]

Answer 17

plugin n = 0,1,2,... and solve x

Answer 18

why is it -13.29

Answer 19

the short answer is :

cos(x) = cos(-x)

Answer 20

there fore, below two equations are one and same :

cos(x) = 2
cos(-x) = 2

Answer 21

you may verify your answer with wolfram :
http://www.wolframalpha.com/input/?i=solve+3cos%282x%29-4sin%282x%29%3D2%2C++0%3Cx%3C2pi

Answer 22

you will need to covnert the solutions to degrees ^

Answer 23

Thank you @ganeshie8

Answer 24

np :)