Question

Solve each equation. Give all solutions in radians in the interval 0 < x < 2pi.
-
a) 2sin x + 3 = 2
b) cosx + 1= 0
c) 3tan^2x = 1
d) 2sin^2x + sinx = 1
e) tanx cosx - tanx = 0

Answer 1

way too many questions. sorry.

Answer 2

Give all solutions in radians in the interval \[0\le x < 2\pi \]

Answer 3

a) \[2\sin x + 3 = 2\]

Answer 4

2 sin x = -1
sin x = -1/2
x = 7pi/6 and 11pi/12

Answer 5

You are probably wondering where I get 7pi/6 and 11pi/12?

Answer 6

yes

Answer 7

Watch.......we know that sin (30 degrees or pi/6) = 1/2
The sine is negative in quadrants 3 and 4.
So if we want to know where sin x = -1/2..we know it will be - in quadrants 3 and 4.

In quadrant 3, we just take 180 + 30 = 210 = 7pi/6
In quadrant 4, we do 360 - 30 = 330 = 11pi/12

Answer 8

response??

Answer 9

is it not 11pi /6

Answer 10

sorry, you are correct, 330 degrees = 11pi/6. I stand corrected!

Answer 11

7pi/6 and 11pi/6

Answer 12

I tried to give you an explanation "short and sweet"...but there's an entire chapter in your text about this.

Answer 13

so is all that my answer?

Answer 14

there are 2 solutions for x, 7pi/6 and 11pi/6.

Answer 15

7pi/6 and 11pi/6 that is my answer correct?

Answer 16

yes.

Answer 17

Your welcome.

Answer 18

b)
cosx + 1 = 0

Answer 19

@Easyaspi314