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WorksheetsSolving Trig Equations - Extra Time
Total questions: 20
Worksheet time: 3hrs 30mins
Why doesn't 2cosx − 3 = 0 have solutions?
cos x is never bigger than one
cos x is never equal to a fraction
Actually, this equation does have a solution, x = π
This equation will have a solution tomorrow.
Which of these is equivalent to 2cos2x − 3cosx = 0 ?
-cos2x = 0
cosx(2cosx + 3) = 0
cosx(2cosx − 3) = 0
cos x = ⅔
Choose a good way to start solving this equation
tanx sin2x = 2 tanx
divide tanx from both sides
factor out tanx
subtract 2 tanx from the left and then factor tanx out
cancel sin2x out
Factor 0 = sin2x − sinx
0 = sinx(sinx)
0 = sinx(1 − sinx)
0 = cosx(sinx − 1)
0 = sinx(sinx − 1)
Factor: sec2x − secx − 2
(sec x)(secx − 2)
(secx − 2)(secx − 1)
(secx − 2)(secx + 1)
(secx + 2)(secx − 1)
To solve the equation sec2x − secx = 2
start by subtracting 2 from both sides and then factor
add sec x to both sides and factor
factor out secx and set each factor equal to 2
sec x is never bigger than 1, so there are no solutions
csc2x = 2 is equivalent to sin2x = ½
True
False
Solve equation for 0≤θ<2π .
1=5−8cosθ
θ=6π,3π,611π
θ=6π,611π
θ=6π
θ=3π,35π
Solve for all values of x over the interval [0,2π]
6π and 67π
65π and 611π
3π and 35π
32π and 34π
Solve equation for 0≤θ<2π .
−sinθcscθ+3cscθ=2sinθ+3cscθ
θ=45π,611π
θ=43π,67π,47π
θ=0,π,45π,47π
θ=45π,47π
Solve equation for 0≤θ<2π .
3sin2θ=7sin2θ+4sinθ+1
θ=67π
θ=0,3π,35π
θ=43π,47π
θ=67π,611π
Solve equation for 0≤θ<2π .
sec2θ+3=2secθ+2
θ=0,2π,32π,35π
θ=3π,π,35π
θ=4π
θ=0
Solve for all values of x over the interval [0,2π]
43πand 45π
4πand 43π
4π,43π,45π and 47π
6π and 65π
Solve equation for 0≤θ<2π .
−2+cotθ=−3
θ=43π,34π,47π
θ=43π,47π
θ=3π,34π
θ=43π,34π
Solve equation for 0≤θ<2π .
2+3cscθ+2csc2θ=csc2θ
θ=23π,611π
θ=43π,47π
θ=67π,23π,611π
θ=4π,611π
Solve equation for 0≤θ<2π .
−1−2sec2θ=−3sec2θ
θ=0,π,34π
θ=0
θ=4π,43π,45π,47π
θ=0,π
2cos2x−3cos(x)+1=0 Solve on the interval [0,2π]
x=0,3π,35π,2π
x=6π,2π,611
x=0,π,2π
x=6π,2π,65π
4sin2x = 3
Find all solutions on the interval [0, 2π) for sec2x − 1=0
x=0, x=π
x=2π, x=23π
x=4π, x=43π
x=0
Solve for all values of x over the interval [0,2π] : tan2 x −2 tanx =−1
4π
4π and 43π
4π and 45π
4π and 47π
