2 of of 2 1 of 3 1 of 1 1 of 2 2 of 2 1 of 1 2 of 2 Diana Ha

2 of of 2 1 of 3 1 of 1 1 of 2 2 of 2 1 of 1 2 of 2 Diana Hanselman 3/1016 9 42 PM Test: M4 Exam (Sec 7.1-7.5) Overview This Question: 5 pts This Test: 100 pts 17 of 20 compiete Time Remaining: 01:12:38 Find all solutions in the interval [0,2x) (T pe an exact answer, using as needed. Use a co mma to separate answers as needed. Use tegers or fractions for any n n en in he 2 of 2 1 of 1 2 of 2 0 of 2 0 of 1 1 of 2 I of 2 1 of 1 0 of 2 O of 2 0 of 1 0 of 2 60min0 of 2 120min 0 of 1 Enter your answer in the answer box Previous Question Next Question Submit Test ) ) 120min 0 of 1

Solution

cos2x -2^1/2*sinx = 1

1-2*sin²x - 2^1/2*sinx - 1 = 0 [ cos2x = 1-2*sin²x]

- ( 2*sin²x+ 2^1/2*sinx) = 0

sinx ( 2*sinx + 2^1/2) = 0

sinx = 0 2sinx + 2^1/2 = 0

sinx = sin(n*pie) 2*sinx = - 2^1/2

x = n*pie n = 0,1,2,3----- sinx = - 1 / 2^1/2

x = 0 , pie ( according to interval given) sinx = -sin(pie/4)

sinx = sin(pie+pie/4) or sin( 2pie - pie/4)

sinx = sin(5pie / 4) or sin (7pie/4)

x = 5*pie/4 or 7pie/4

ans = x = 0,pie, 5*pie/4, 7*pie/4