Please help with trig equation!?

Question

6sin^3x - 4sin^2x - 3sinx +2 =0 all solutions, in radians

Answer 1

First we make a substitution of variables: Let z=sin x. Then we have:

6z³ - 4z² - 3z + 2 = 0

Possible rational solutions are ±2, ±1, ±2/3, ±1/2, ±1/3, and ±1/6. Brute force search reveals that 2/3 is such a rational solution, so to find the other two, we employ synthetic division:

2/3| 6..... -4..... -3..... 2
............... 4...... 0.... -2
-------------------------------
...... 6...... 0..... -3..... 0

so the equation factors as (z-2/3) (6z²-3). The second expression can in turn be solved for its roots:

6z²-3=0
z²-1/2=0
z²=1/2
z=±1/√2

So we have three solutions: z=2/3, z=1/√2, and z=-1/√2. But z=sin x, so we now set sin x equal to each of these and find all of the solutions:

sin x=1/√2 → x=π/4±2πk or x=3/4±2πk
sin x=1/√2 → x=-π/4±2πk or x=-3π/4±2πk

These previous four sets of solutions may be combined into a single expression: x=π/4±kπ/2.

You also have the possible solution:

sin x=2/3

Since 2/3 is not the sine of any "nice" angle, you simply write the solutions here as
x=arcsin (2/3)±2πk or x=π-arcsin (2/3)±2πk

Thus your entire solution set is:

x=π/4±kπ/2 or x=arcsin (2/3)±2πk or x=π-arcsin (2/3)±2πk.

Answer 2

for simplicity and to avoid any confusions in the readability while solving let sinx=a

so we have
6a^3-4a^2-3a+2=0
the common term in the first two terms is 2a^2
2a^2(3a-2)-1(3a-2)
(2a^2-1)(3a-2)
substituting for a=sin x we have the answer is
(2sinx^2-1)(3sinx-2)

Answer 3

If you substitute Sin x for y, you get the solvable equation 6y^3-4y^2-3y+2. Factoring, we get:
2y^2(3y-2) - 1(3y-2)

(3y-2)(2y^2-1)

solving for y, we get:

y= 2/3
y^2 = 1/2
which simplifies to:
y = 1/4
y = -1/4

Then bring back the sin x to get
sin x = 2/3
sin x = 1/4
sin x = -1/4

x = .729, .25268, 6.030505052