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, 360 – 75.522  75.5, 284.5 Hence x = 0, 75.5, 180, 284.5, 360. 1 2 1 2 1 2 cos x = - 2 3 Reference angle = cos−1 2 3  48.190 Since cos x < 0, x is in the 2nd or 3rd quadrant. x  180 − 48.190, 180 + 48.190  131.8, 228.2

4 (b) 2cos x + cot x = 0 2cos x + cos x sin x = 0 cos x (2 + 1 sin x ) = 0 cos x = 0 or sin x = - 1 2 cos x = 0 x = 90, 270 sin x = - 1 2 Reference angle = sin−1 1 2 = 30 Since sin x < 0, x is in the 3rd or 4th quadrant. x = 180 + 30, 360 − 30 = 210, 330 Hence x = 90, 210, 270, 330. 4 1cos(2 30 ) 2 Reference angle 60 Take note on range: Since 0 360 , 0 2 720 , 30 2 30 750 2 30 180 60 , 180 60 , 540 60 , 540 60 2 30 120 , 240 , 480 , 600 75 , 105 , y y y y y y y +  =− =        +   + = −  +  −  +  + =     =   225 , 285 5 place) decimal 1 o(correct t 2.348 ,8.281 ,168.2 ,8.101 422.696 ,578.563 ,422.336 ,578.2032 578.23720 ,578.23540 ,578.23360 ,578.231802 places) decimal 3 o(correct t 578.23 5 22sin 022sin5 = = −+−+= = −= =+ x x x x x 

5 6 place) decimal 1 o(correct t 9.334 ,9.224 ,9.154 ,9.64 806.669 ,806.489 ,806.309 ,806.1292 194.50720 ,194.50540 ,194.50360 ,194.501802 places) decimal 3 o(correct t 194.50 5 62tan = = −−−−= = −= y y y y  7 fig.) sig. 3 o(correct t 77.4 ,26.4 ,63.1 ,12.1 0654.14 ,0654.13 ,0654.12 ,0654.18 52 fig.) sig. 5 o(correct t 0654.1 8 7 8 52sin 4 7 8 52sin2 = −+−+=+ = −=     + −=     + x x x x     Since 5.41  x , 4.26 1.63, ,12.1=x . 8 Hence, 2 3 ,46.3 ,2 ,322.0 =x . 9 (a) 2 sin x – 5 cosec x = 9 2 sin x – 5 sin x = 9 2 sin2 x – 9 sin x – 5 = 0 (2 sin x + 1)(sin x – 5) = 0 sin x = – 1 2 or sin x = 5 (no solution) Reference angle = sin−1 1 2 = p 6 Since sin x < 0, x is in the 3rd or 4th quadrant. x = p + p 6 , 2p - p 6 0)sin3(coscos 0cossin3cos cossin3cos 2 2 =− =− = xxx xxx xxx 2 3 ,2 0cos = = x x or fig.) sig. 3 o(correct t .463 ,322.0 32175.0 ,32175.0 fig.) sig. 3 o(correct t 32175.0 3 1tan sin3cos 0sin3cos = += = = = =− x x x xx xx  

6 = 7p 6 , 11p 6 (b) 14 sin x + 54 cos x = 18 sin x + 27 cos x 4sin x = 27cos x sin x cos x = 27 4 tan x = 27 4 Reference angle = tan−1 27 4  1.424 Since tan x > 0, x is in the 1st or 3rd quadrant. x  1.424, π + 1.424  1.42, 4.57 10 ( ) 0)1cos6(cos 0coscos6 3coscos63 3coscos213 3cos2cos3 2 2 2 =− =− −=+− −=+− −= yy yy yy yy yy Either = = 270 ,90 0cos y y Or place) decimal 1 o(correct t 6.279 ,4.80 406.80360 ,406.80 406.80 6 1cos 1cos6 01cos6 = −= = = = =− y y y y y  Hence, = 6.279 ,270 ,90 ,4.80y .

7 11 44.5 ,99.3 84806.02 ,84806.0 57.1 84806.0 2 4 3sin Basic 1sinor 4 3sin 0)1)(sin3sin4( 03sinsin4 02sin2)sin21(4 2sin22cos4 1 2 2 = −+= == =    = =−= =−+ =−− =++− −=+ −   x x xx xx xx xx xx 12 (Rej.) 2 1cos 5cos 2 secor 5 1 sec 0)2 sec)(1 sec5( 02sec 11sec 5 sec 117)1sec(5 sec 117 tan5 2 2 2 == == =−− =+− =+− =+ yy yy yy yy yy yy 13 0)1)(cos1cos3( 01cos2cos3 0cos22cos4cos1 2 22 =−+ =−− =−−+− yy yy yyy = == =−= 5.250 ,5.109 360 ,0 53.70 Basic 1cos or 3 1cos y y yy