Approximate ............................

Question

Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°). (If there is no solution, enter NO SOLUTION.)

 

sin² t − 6 sin t + 2 = 0

 

t	=	°	'	(smaller value)
t	=	°	'	(larger value)

Answer 1

If we let x = sin t, the equation becomes:

 

x² - 6x + 2 = 0

 

Use the quadratic formula to solve:

 

x = (6/2)±(1/2)√((-6)²-4(1)(2))

x = 3 ± √7

x = sin(t) ≅ 0.354 and 5.646

 

Since the range of sin(t) is ±1:

 

sin(t) = 0.354

t = sin^-1(0.354) ≅ 20^o45' and 159^o15'