Question

solve the equation for x:

1 - sin(x) = (root3)(cos(x))

2sin^3(x)+sin^2(x)-2sinx=1

Answer 1

just a start would be incredible

Answer 2

are these two separate problems or a system of equations?

Answer 3

two different problems, forgot the s right after solve the equation, sorry!

Answer 4

I'll help you with the first problem and you can post the second as a different question. By root do you mean square root? (sqrt)

Answer 5

I think i understand the first problem

Answer 6

its the double angle formula that is required, and after that it's pretty straightforward

Answer 7

square both sides in the first problem:

(1 - sin(x))^2 = ((root3)(cos(x)))^2

1 - 2sinx + sin^2 x = 3 cos^2 x

1 - 2sinx + sin^2x = 3(1-sin^2 x)

now it should be easy from here

Answer 8

for my solution set I got:
\[x = \pi/6 + 2\pi n\]\[x=3\pi/2 + 2\pi n\]

Answer 9

4x^2 - 2sinx -2 = 0

2sin^2 - sinx - 1 = 0
(2sinx + 1)(sin x - 1) = 0

sin x = -1/2
sin x = 1


so solution set:

x = -pi/6 + 2npi
or
x = pi/2 + 2npi

Answer 10

doesn't 7pi/6 also make sinx=-1/2?

Answer 11

So then there would be three solution sets if what I said above is true.

Answer 12

because we squared the equations, all the solution sets won't work, because some will give a positive RHS while the LHS gives a negative, and vice versa. Consider 7pi/6:

LHS: 1 - (-1/2) = 3/2
RHS: sqrt3 * -sqrt3/2 = -3/2

Therefore RHS and LHS are not equal

Answer 13

I see, so then it is only be pi/2 and 11pi/6?

Answer 14

if the solution needs to be in [0, 2pi], yes

Answer 15

awesome! So for the second problem, I took out the sinx and made it 2sin^x+sinx-2, all of that being equal to 1. Does that sound like the right way to start it?

Answer 16

2sin^3(x)+sin^2(x)-2sinx=1

for this one, it's important to realize you can use a group factor method:

Rewrite it like this:

2sin^3(x)+sin^2(x)-2sinx -1 = 0

you can see absolute value of the coefficients, in order of powers (rank), goes:

2 1 2 1

Now you can do this:

sin^2 x(2sin x + 1) - (2sin x + 1) = 0

-> (2sin x + 1)(sin^2 x - 1) = (2sin x + 1)(sin x + 1)(sin x - 1) = 0

now you can solve from there

Answer 17

I found 4 solution sets:
\[3\pi/2 + 2\pi n\]\[\pi/2 + 2\pi n\]\[11\pi/6+2\pi n\]\[7\pi/6 + 2\pi n\]

Answer 18

Checked them all as well, none are extraneous