Question

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

Answer 1

`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^@`

Answer 2

`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]}{