# Solve for x in (0,pi) : 2sin xcos x - 3 - 6 sin x + cos x = 0Solve for x in (0,pi) : 2sin xcos x - 3 - 6 sin x + cos x = 0

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to group the members of equation such that:

`(2sin xcos x + cos x) - (3 + 6 sin x) = 0`

Factoring out `cos x` in the first group and 3 in the second group yields:

`cos x(2sin x + 1) - 3(1 + 2sin x) = 0`

Factoring out `(1 + 2sin x)` yields:

`(1 + 2sin x)(cos x - 3) = 0`

Using zero product property yields:

`{(1 + 2sin x = 0),(cos x - 3 = 0):} => {(2sin x = -1),(cos x=3):}`

`sin x = -1/2` invalid if `0<= x <= pi`

`cos x = 3` invalid since `cos x in [-1,1]`

Hence, evaluating the solution to the equation, under the given conditions, yields that there exists no x values in `[0,pi]` for the equation holds.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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2sin x*cos x - 3 - 6 sin x + cosx = 0

We could  factorize the first term and the3rd term, by cos t:

cos x(2sin x+1) - 6 sin x - 3 = 0

We'll also factorize the last 2 terms by -3:

cos x(2sin x+1) - 3(2sin x + 1) = 0

We'll factorize again by 2sin x+1:

(2sin x+1)(cos x - 3) = 0

We'll set the first factor as zero:

2sin x + 1 = 0

We'll subtract 1:

2sin x = -1

sin x = -1/2

x = arcsin (-1/2)

Since we have to solve the equation in the range (0,pi), we'll validate the solution from the unit circle.

The sine function is positive in the range (0,pi), so there is no solutions for sin x = -1/2, over the range (0,pi)

We'll set the 2nd factor as zero:

cos x - 3 = 0