Find the solution(s) of the equation in the interval [0, 2π). Use a graphing utility to verify your results. (Enter your answers as a comma-separated list.) 16 sin x + π 2 + 24 tan(π − x) = 0

Answer 1

sin(x+ pi/2) = sin x cos(pi/2) + cos x sin(pi/2)

= sin x * 0 + cos x * 1

= cos x

tan ( pi - x) = (tan pi - tan x) / (1 + tan pi * tan x)

= ( 0 - tan x)/ ( 1 + 0 * tan x)

= -tan x/ 1 = -tan x

the equation is :

16 * cos x - 24 tan x = 0

16 * cos^2 x - 24 * sin x = 0 <--- multiplies by cosine

16 ( 1 - sin^2) - 24 sin x = 0 <-- substitutes cos^2 = 1 - sin^2 from trig identity; argument dropped

16 - 16 sin ^2 - 24 sin = 0 <--- distributes the 16

16 sin ^2 + 24 sin - 16 = 0 <--- multiplies both sides by -1

2 z ^2 + 3 z - 2 = 0 <-- substitutes z = sin and divides both sides by 8

( 2z - 1 )(z + 2 ) = 0

2z - 1 = 0 or z + 2 = 0

z+2=0 --> z = -2 and the sin can never be that small

z = 1/2 is the only solution

sin x = 1/2

sine is positive in quadrants 1 and 2.

x = 30 and 150 which in radians is x=pi/6 and 5*pi/6