Q1.Solve for 0<X<2pi and then give the general solution:i) tan2X= -1 (ii) sin2X= -1 (iii) sin 2X = 0.8910

Question

solve the following:(i) 4 sin2X( sinX-1)=0
ii) sin2X =-1

Answer 1

1)

tan2X = -1
2tanx / (1 - tan(^2)x) = -1
2tanx = -1 + tan(^2)x
tan(^2)x - 2tanx -1 = 0
This can be factored as:
(tanx - 2) (tanx + 1)
Therefore tanx = 2, or tanx = -1
for tanx = 2:
using your calculator,
x = 1.1071 radians
Adding π for the other solution gives you 4.2487 radians.

for tanx = -1
using your calculator,
x = -0.7854
You don't want this solution as you want everything between 0 and 2π.
So your solutions will be π - 0.7854 and 2π - 0.7854
x = 2.3562 radians or x = 5.4878 raidans

Therefore x = 1.1071, 2.3562, 4.22487, or 5.4878 radians

2)

For the second question, sin2x = -1

Thefore 2x = 3π/2
Therefore
x = (3π/2) / 2
x = 3π / 4 OR x = (3π / 4) + (2pi/2)
x = 3π / 4 OR x = 7π / 4

3)

Now for sin2x = 0.8910
Using your calculator, 2x = 1.09954 radians,
therefore x = 0.5498 radians.
The other solutions can be found by adding π/2, 2(π/2), 3(π/2) to 0.5498
therefore x = 0.5498, 2.1296, 3.6914, or 5.2622 radians.

4)

4 sin2x (sin x - 1) = 0
4 (2sinxcosx) (sinx - 1) = 0
8sinxcosx (sinx - 1) = 0
sinxcosx (sinx - 1) = 0/8

sinxcosx = 0 OR sin x - 1 = 0
Therefore
either sin x = 0 or cos x = 0 or sin x = 1

Dealing with sin x = 0

x = 0, π, 2π

cos x = 0
x = π/2, 3π/2

sin x = 1
x = π/2

Therefore x = 0, π/2, π, 3π/2, or 2π

your last question is a repeat of a question already answered in your subject.

Answer 2

2x = arctan(-1)

that is in quad 2 and 4
arctan 1 =pi/4

2x = 3pi/4 and 7pi/4
x = 3pi/8 and 7pi/8
gen sol = 3pi/8 + 2npi or 7pi/8 + 2npi

sin2x = -1
2x = arcsin -1 = 3pi/2
x = 3pi/4
gen sol = 3pi/4 + 2npi

the last one you need to find arcsin using a calc.

Answer 3

i)

tan(2x) = -1

2x = 3pi/4, 7pi/4, 11pi/4, 15pi/4

x = 3pi/8, 7pi/8, 11pi/8, 15pi/8

ii)

sin(2x) = -1

2x = 3pi/2, 7pi/2

x = 3pi/4, 7pi/4

iii)

sin(2x) = 0.8910

2x = sin^-1(0.8910)

2x = 7pi/20, 13pi/20, 47pi/20, 53pi/20

x = 7pi/40, 13pi/40, 47pi/40, 53pi/40

1v)

4 sin(2x)(sinx - 1) = 0

either sin(2x) = 0 or sinx = 1

2x = pi, 2pi, 3pi

x = pi/2, pi, 3pi/2

v)

sin(2x) = -1

same as ii problem

Answer 4

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