How do you solve #2\sin x = 3\cos ( x + 45^ { \circ } )#?

2 Answers
Jul 27, 2018

#color(brown)(x = arctan (3 / (3 + 2 sqrt2) ) ~~ 27.2357^@#

Explanation:

#2 sin x = 3 cos (x + 45^@)#

#color(crimson)("Using identity " cos (A + B) = cos A cos B - sin A sin B,#

#2/3 sin x = cos x * cos 45 - sin x * sin 45#

#2/3 sin x = cos x / sqrt 2 - sin x / sqrt2#

#((2 sqrt 2)/3) sin x = cos x - sin x#

#cos x / sin x - (cancel sin x / cancel sin x )^color(red)(1)= (2 sqrt2) / 3#

#cot x = 1 + (2 sqrt2) / 3#

#tan x = 1 / (1 + ((2 sqrt2)/3))#

#tan x = 3 / (3 + 2 sqrt2) #

#x = arctan (3 / (3 + 2 sqrt2) )#

#color(brown)(x ~~ 27.2357^@#

Jul 27, 2018

#x = ( kpi + 0.47535)) rad = (180 k + 27.2357)^o#,
# k = 0, +-1, +-2, +-3, ...#

Explanation:

#= 2 sin x - 3 cos ( x + pi/4 ) #

#= 2 sin x - 3 (cos x cos (pi/4) - sin x sin (pi/4) )#

#= ( ( 2 +3/sqrt2 )sin x - 3/sqrt2 cos x)#

# = a ( sin x sin alpha - cos x cos alpha )#

#= a sin ( x - alpha )#,

where

#a = sqrt(( 2 + 3/sqrt2 )^2 + ( 3/sqrt2)^2 )#

# = sqrt(13 +6sqrt2 ) = 4.63522#, nearly.

#cos alpha = ( 2 - 1/sqrt2 )/a and sin alpha = (3/sqrt2)/a#. Now,

sin ( x - alpha ) = 0, and so,

#x - alpha = kpi, k = 0, +-1, +-2, +-3,...#, giving

#x = kpi + alpha = (180 k + arcsin (0.45765))^o#

#= (180 k + 27.2357))^o = ( kpi + 0.47535)) rad#.

GRAPH CHECK, 6-sd x = 0.474353 rad :
graph{y-2 sin x + 3 cos ( x + pi/4 ) = 0[0.47535.475356 -0.00001 0.00001]}{