Trig Equations with a Twist

Trig Equations with a Twist
Solve each equation on the interval 0 ≀ π‘₯ < 2πœ‹ with exact values.
_________________ 1.
2 sin2 π‘₯ βˆ’ 5 sin π‘₯ + 2 = 0
_________________ 2.
sin2 π‘₯ βˆ’ 2 sin π‘₯ βˆ’ 3 = 0
_________________ 3.
4 sin2 π‘₯ βˆ’ 3 = 0
Use identities to assist you in solving each of the following on the interval 0 ≀ π‘₯ < 2πœ‹ with exact values.
_________________ 4.
2 sin π‘₯ = csc π‘₯
_________________ 5.
1 βˆ’ 3 cos π‘₯ = sin2 π‘₯
_________________ 6.
tan2 π‘₯ = 2 sec π‘₯ βˆ’ 1
_________________ 7.
cot 2 π‘₯ = 3(csc π‘₯ βˆ’ 1)
_________________ 8.
2 cos 2 π‘₯ + sin π‘₯ = 1
_________________ 9.
2 cos 2 π‘₯ βˆ’ √3 cos π‘₯ = 0
_________________ 10. cos 2π‘₯ + 3 cos π‘₯ βˆ’ 1 = 0
_________________ 11. sin 2π‘₯ = βˆ’ sin π‘₯
_________________ 12. tan 2π‘₯ = cot π‘₯
Consider the period and/or frequency of the graphs when solving each of the following on the interval
0 ≀ π‘₯ < 2πœ‹ with exact values.
1
_________________ 13. sin 2π‘₯ = 2
_________________ 14. cos 3π‘₯ =
√2
2
_________________ 15. sin 3π‘₯ = βˆ’
√3
2
Trig Equations with a Twist KEY
Solve each equation on the interval 0 ≀ π‘₯ < 2πœ‹ with exact values.
πœ‹ 5πœ‹
π‘₯= ,
6 6
_________________
1. 2 sin2 π‘₯ βˆ’ 5 sin π‘₯ + 2 = 0
(2 sin π‘₯ βˆ’ 1)(sin π‘₯ βˆ’ 2) = 0
2 sin π‘₯ βˆ’ 1 = 0 or sin π‘₯ βˆ’ 2 = 0
3πœ‹
π‘₯=
2
_________________ 2.
1
sin π‘₯ = 2 or sin π‘₯ = 2
sin2 π‘₯ βˆ’ 2 sin π‘₯ βˆ’ 3 = 0
(sin π‘₯ + 1)(sin π‘₯ βˆ’ 3) = 0
sin π‘₯ + 1 = 0 or sin π‘₯ βˆ’ 3 = 0
πœ‹ 2πœ‹ 4πœ‹ 5πœ‹
π‘₯= , ,
,
3 3 3 3
_________________
3.
sin π‘₯ = βˆ’1 or sin π‘₯ = 3
4 sin2 π‘₯ βˆ’ 3 = 0
4 sin2 π‘₯ = 3
3
sin2 π‘₯ = 4
sin π‘₯ = ±
√3
2
Use identities to assist you in solving each of the following on the interval 0 ≀ π‘₯ < 2πœ‹ with exact values.
πœ‹ 3πœ‹ 5πœ‹ 7πœ‹
π‘₯= , ,
,
𝟏
4 4 4 4
_________________
4. 2 sin π‘₯ = csc π‘₯
2 sin π‘₯ =
𝐬𝐒𝐧 𝒙
2 sin2 π‘₯ = 1
1
sin2 π‘₯ = 2
sin π‘₯ = ±
√2
2
πœ‹ 3πœ‹
π‘₯= ,
2 2
_________________
5.
1 βˆ’ 3 cos π‘₯ = sin2 π‘₯
πœ‹ 5πœ‹
π‘₯= ,
3 3
_________________ 6.
1 βˆ’ 3 cos π‘₯ = 𝟏 βˆ’ 𝐜𝐨𝐬𝟐 𝒙
0 = 2 cos 2 π‘₯
0 = cos 2 π‘₯
tan2 π‘₯ = 2 sec π‘₯ βˆ’ 1
𝐬𝐞𝐜 𝟐 𝒙 βˆ’ 𝟏 = 2 sec π‘₯ βˆ’ 1
sec 2 π‘₯ βˆ’ 2 sec π‘₯ = 0
sec π‘₯ (sec π‘₯ βˆ’ 2) = 0
sec π‘₯ = 0 or sec π‘₯ = 2
πœ‹ πœ‹ 5πœ‹
π‘₯= , ,
2 6 6
_________________
7.
cos π‘₯ = βˆ… or cos π‘₯ =
cot 2 π‘₯ = 3(csc π‘₯ βˆ’ 1)
1
2
csc 2 π‘₯ βˆ’ 1 = 3(csc π‘₯ βˆ’ 1)
csc 2 π‘₯ βˆ’ 1 = 3 csc π‘₯ βˆ’ 3
csc 2 π‘₯ βˆ’ 3 csc π‘₯ + 2 = 0
(csc π‘₯ βˆ’ 1)(csc π‘₯ βˆ’ 2) = 0
csc π‘₯ = 1 or csc π‘₯ = 2
1
sin π‘₯ = 1 or sin π‘₯ = 2
πœ‹ 7πœ‹ 11πœ‹
π‘₯= ,
,
2 6 6
_________________
8.
2 cos 2 π‘₯ + sin π‘₯ = 1
2(𝟏 βˆ’ 𝐬𝐒𝐧𝟐 𝒙) + sin π‘₯ = 1
2 βˆ’ 2 sin2 π‘₯ + sin π‘₯ = 1
2 sin2 π‘₯ βˆ’ sin π‘₯ βˆ’ 1 = 0
(2 sin π‘₯ + 1)(sin π‘₯ βˆ’ 1) = 0
1
πœ‹ 3πœ‹ πœ‹ 11πœ‹
π‘₯= , , ,
2 2 6 6
_________________
9.
sin π‘₯ = βˆ’ 2 or sin π‘₯ = 1
2 cos 2 π‘₯ βˆ’ √3 cos π‘₯ = 0
cos π‘₯ (2 cos π‘₯ βˆ’ √3) = 0
cos π‘₯ = 0 or cos π‘₯ =
πœ‹ 5πœ‹
π‘₯= ,
3 3
_________________
10. cos 2π‘₯ + 3 cos π‘₯ βˆ’ 1 = 0
√3
2
(𝟐 𝐜𝐨𝐬𝟐 𝒙 βˆ’ 𝟏) + 3 cos π‘₯ βˆ’ 1 = 0
2 cos 2 π‘₯ + 3 cos π‘₯ βˆ’ 2 = 0
(2 cos π‘₯ βˆ’ 1)(cos π‘₯ + 2) = 0
1
cos π‘₯ =
2πœ‹ 4πœ‹
π‘₯ = 0, πœ‹,
,
3 3
_________________
11. sin 2π‘₯ = βˆ’ sin π‘₯
or cos π‘₯ = βˆ’2
2
2 sin π‘₯ cos π‘₯ = βˆ’ sin π‘₯
2 sin π‘₯ cos π‘₯ + sin π‘₯ = 0
sin π‘₯ (2 cos π‘₯ + 1) = 0
1
sin π‘₯ = 0 or cos π‘₯ = βˆ’ 2
πœ‹ 5πœ‹ 7πœ‹ 11πœ‹
π‘₯= ,
, ,
6 6 6 6
_________________
12. tan 2π‘₯ = cot π‘₯
𝟐𝐭𝐚𝐧 𝒙
𝟏
πŸβˆ’π­πšπ§πŸ 𝒙
= 𝐭𝐚𝐧 𝒙 cross mult.
2 tan2 π‘₯ = 1 βˆ’ tan2 π‘₯
3 tan2 π‘₯ = 1
1
tan2 π‘₯ =
3
1
tan π‘₯ = ±βˆš3 = ±
√3
3
Consider the period and/or frequency of the graphs when solving each of the following on the interval
0 ≀ π‘₯ < 2πœ‹ with exact values.
πœ‹ 5πœ‹ 13πœ‹ 17πœ‹
π‘₯=
,
,
,
1
πœ‹ 5πœ‹
2πœ‹
12πœ‹
12 12 12 12 13. sin 2π‘₯ =
_________________
2π‘₯ = 6 , 6
𝑇 = 𝟐 = 12
2
πœ‹
5πœ‹
π‘₯ = 12 , 12 then add
πœ‹ 7πœ‹ 9πœ‹ 15πœ‹
π‘₯=
, ,
,
,
12 12 12 12 14. cos 3π‘₯ = √2
_________________
2
17πœ‹ 23πœ‹
,
12 12
3π‘₯ = 4 ,
4πœ‹ 5πœ‹ 10πœ‹ 11πœ‹
π‘₯=
,
,
,
,
√3
9 9 9
9
_________________
15. sin 3π‘₯ = βˆ’ 2
16πœ‹ 17πœ‹
,
9
9
3π‘₯ =
12πœ‹
12
πœ‹ 7πœ‹
πœ‹
to each till you reach 2πœ‹
𝑇=
4
7πœ‹
2πœ‹
πŸ‘
=
8πœ‹
12
8πœ‹
π‘₯ = 12 , 12 then add 12 to each till you reach 2πœ‹
π‘₯=
4πœ‹ 5πœ‹
3
,
3
4πœ‹ 5πœ‹
9
,
9
𝑇=
then add
6πœ‹
9
2πœ‹
πŸ‘
=
6πœ‹
9
to each till you reach 2πœ‹