XMS Trigo equation practice

1 Xinmin Secondary School Additional Mathematics Practice Topic: Solving Trigo Equations Name: ______________________________ ( ) Date: ___________ 1 Given that 0 < x < 360, solve the equation (a) sin x = - 1 3 , (b) cosec x = tan 60. 2 Given that 0 ≤ x ≤ 360, solve the equation (a) 2 cos2 x – 5 cos x – 3 = 0. (b) 6 sin2 x = 2 – 2 cos x. 3 Given that 0 ≤ x ≤ 360, solve the equation (a) tan x = 4 sin x, (b) 2 cos x + cot x = 0. 4 Solve the equation 2 1)302cos( −=+y for all the angles between 0 to 360 inclusive. 5 Solve the equation 022sin5 =+x for  3600 x . Correct your answer to 1 decimal place. 6 Find all the angles between 0 and 2 which satisfy 2 1 62tan =     −x . Correct your answer to 3 significant figures. 7 Solve the equation 4 7 8 52sin2 −=     + x for 5.41  x . Correct your answer to 3 significant figures. 8 Find all the angles between 0 and 2 which satisfy the equation xxx cossin3cos2 = . If the answer is not exact, correct the answer to 3 significant figures. 9 Given that 0 ≤ x ≤ 2 , solve the equation (a) 2 sin x – 5 cosec x = 9, (b) 14 sin x + 54 cos x = 18 sin x + 27 cos x. 10 Solve the equation 3cos2cos3 −= yy for  3600 y . If the answer is not exact, leave the answer to 1 decimal place. 11 Solve the equation 20for 2sin22cos4 −=+ xxx .

2 12 Find all the angles between  360 and 0 which satisfy 5 tan2 y + 7 = 11 sec y. 13 Solve the equation yyy cos22cos2sin 2 =+ for  3600 y . Work Solution 1 (a) sin x = - 1 3 Reference angle = sin−1 1 3  19.471 Since sin x < 0, x is in the 3rd or 4th quadrant. x  180 + 19.471, 360 – 19.471  199.5, 340.5 (b) cosec x = tan 60 1 sin x = 3 sin x = 1 3 Reference angle = sin−1 1 3  35.264 Since sin x > 0, x is in the 1st or 2nd quadrant. x  35.264, 180 − 35.264  35.3, 144.7

3 2 (a) 2cos2 x – 5cos x – 3 = 0 (2 cos x + 1)(cos x – 3) = 0 cos x = – or cos x = 3 (no solution) cos x = – Reference angle = cos−1 = 60 Since cos x < 0, x is in the 2nd or 3rd quadrant. x = 180 − 60, 180 + 60 = 120, 240 (b) 6 sin2 x = 2 – 2 cos x 6(1 – cos2 x) = 2 – 2 cos x 6 – 6 cos2 x = 2 – 2 cos x 3 cos2 x – cos x – 2 = 0 (3 cos x + 2)(cos x – 1) = 0 cos x = - 2 3 or cos x = 1 cos x = 1 x = 0, 360 Hence x = 0, 131.8, 228.2, 360. 3 (a) tan x = 4 sin x sin x cos x = 4 sin x sin x – 4 sin x cos x = 0 sin x (1 – 4 cos x) = 0 sin x = 0 or cos x = 1 4 sin x = 0 x = 0, 180, 360 cos x = 1 4 Reference angle = cos−1 1 4  75.522 Since cos x > 0, x is in the 1st or 4th quadrant. x  75.522, 3