MHF 4U1 Trigonometric Equations 1. Solve for x, Ox2,r. a) sinx=—-2 b) cosx=— 2 c) tanx=—1 d) secx e) sinx 2x = 0 cos = —2 = — — 2. Solveforx,—2rxO. a) cosx b) tan x 2 = d) sin2x= = tanx c) sin 2 x—sinx = 2 2 =9 f) (2cscx—l) x—3=O 2 e) 4cos 3. Solveforx,Ox22T. a) cos(2x) + cos x + 1 b) cos(2x) = sin x c) 3 tan x = tan(2x) g) sin(2x)+ cos(2x)= 0 e) sin(2x)cosx + sin 2x=1 h) 3cos(2x)+ 2 + cosx = 0 sin(2x)+ sinx = 0 i) sin(2x)= tanx j) k) 3sin(2x)+cosx = 0 1) 3sinx+cos(2x)= 2 = 0 d) sinx = 6sin(2x) 2 x—sin cos 2x = 2sinxcosx 4. Solve for x in the given interval. a) sinx—sinxtanx=O, Oxr b) cosxtan(3x)=0, —rx0 c) ósin x—5cosx—2=0, x 2 2 2 e) 2 x—3sin cos x =1, —2nx2n d) sinx+tanx=O, —x 0 2tanx=secx, —2rx0 5. Solve for x in the given interval. a) b) sin(2x)=cosx, Z x cos(2x)=cos x 2 , —r x ,r c) cos 2x — 2sinxcosx — 2x sin = 0, 0 x d) 2(sin x + cos 4 x)= 1, — ,r x r 6. Solve for x in the given interval. a) cosx=, —2x2 b) sin(2x)= c) 2 sin x +sinx=O, x d) 2sin x+sinx—1=0, 0x4 2 e) 2sec x+3secx—2=0, 0x2 2 0 tanx+sec(2x)=1, x x - — 3 f,’ 0 1 l”43 ,. . hi’ 4 j. )\ n tJ Hi HI )c, 0 Ij ct I “4 0 I I,’ ui VJ I. 4 H ‘ii Lj 4% 0 0 0 4 4- 4 I 4 IL 0 A -‘ ‘ cJ — ‘4ft 1% 0 1 t .L - ‘-. (A - I 44 < -- ‘4, - (4 x L - I’ I’ ‘I F’) p 0 ‘$1 >p 1, ‘-4 ‘I Y. ‘1 1- 1• sJ -ó — -A \s 0 C 0 ‘a — I C’ J 0 .4-) 0 o ‘a L —I” 0 0 c 0 c s’ 4% 3 ‘ ( 0 1 — ,L (A ‘ o + c4 41 ( I.. 1* a a a C 0 L , .1 4 (t( I, 4 I’ 0 0 4 0 c 41 0 Lf p It ii—’ c) ( CA Ii 3 0 N 0 — I 4.. 00 (4 4 L 0 4.. 0 0 i S 1 ii I 1’ 0 0 c ,c ‘ ) 3 N ‘I r) v-I -4- c%J 1 k ‘.. (fi 4 (1 7 4’ 0 r\’m I ‘L .4L -p J ‘4’ ‘‘ N 0 0’ 1) ‘ L ‘, .‘j’ •*‘‘ . ‘ ‘ .4. , a ‘4 ‘S . L .. \ .,T 4, ‘,., .1 .4. 0 L. V c:fr I” (C%J 0 0 7 a 0 1’ P -I —I- N I I 0 1’ 0 ,1 7 0 I, tir’ F 0 p r 7’ )_ -% I, ‘1 7c 1 0 P ‘I’ “3 Co (, S - I 4%J t t L. or I 3(111 -O 0 C) (rt’)e or 0 X: t’’() :r >: D. )L c)r Qf iii St’’ t 7,.3o ,) o ‘. — I) — 0 7 0 11 — I 1, ‘7S 7, ‘1 0 — 0 0 i4 1 4.-, ‘I I 7- ‘I) ‘p ‘I, 0 S. 1 7; “1 V I, 4- ‘JJ U’ ‘I’ tJ I + 0f ,€—. 4- ‘0 4( 0 i, 0 1 0 00 /N o I .0 I. 1 N t1 o c. 0 - 0 0 1 c 0 c 0 A S 0 -J 1 1 1 c. j .4 ‘4 L 3 3 ( 0 4 ‘F’ ii 0 0 “1 c 1r a & 0 1 ‘-I C,’ I 0 0’ I Cd ‘I’ 1 0 4 (I’ L. L. 0 I L L 1 0 (%J 0 —4,, 1 S 1 1’ I 0 ,- I 1. (4 .f g a1. 0 a & 4 4JS ‘4’ ‘ Jb x 0 0 C’ IL L 0 iiJ 4 V. ‘\i L ¶1’ ‘4’ c:r 0 I’ t 43 —-——--— “ ‘ —‘ LJ i’ ‘— ‘ .4 4%) e C44 %! 4’ 41 ci 0 -J 3 4, 0 ‘1 1 1’ ‘1 9:1 1, s 1 ; .4 : )c \-J \ * H- “ 0 - , — w-. %%J 1, 1. ç ‘ 1 __ ,.-, ‘ii I’ ‘I ‘4’ ‘4 ( 0 L 0 L1%9 C’ I L 0 4’ 0 ‘a i:I%J ‘.- .,“ 4- Li vi ‘%_ — ( 0’ a vi C’)) I a 4’ L 0 1 I Ø1 4- (i’ ( p ‘U. ( I’ 1 C,, -4 41 + 3’ :‘D tcJ ‘I. U C’ - 4 _ir -1 ‘-I 0 4- 1 Cd 44 6 L -4--- ‘7, LI, q) ‘I ‘I, 4 . F c “3 0 .1 1 13 v) ,Th -,_ 3: - -- —V — , Li) I-) 0 C’ ( -w ‘1 0 r x ‘I, 1 7, x 7’, ‘/1 . I 4. I 31 9 LI I 11 1 7. C’ 0 ÷ 7L ‘0 (-I, .. (4 LA C-’ ‘4 *1 ‘4 ‘4 I —& .4 ‘1 ‘3 (A t1 44 ‘-S .4- 4 C a x t I 0 •1 C’ b c’1 ‘3’ ‘9 ii cJ 4- .7’ 4-” 0 p. I (1’ 4- r.c3 0 V.’ I ti 0 14. (. I ‘4
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