Find all solutions to the equation sin2x+sinx-2cosx-1=0 in the interval [0, 2pi)

The solutions appear to be {π/2, 2π/3, 4π/3}.

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Replacing sin(2x) with 2sin(x)cos(x), you have
  2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
  sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
  (sin(x) -1)(2cos(x) +1) = 0

This has solutions
  sin(x) = 1
  x = π/2
and
  2cos(x) = -1
  cos(x) = -1/2
  x = {2π/3, 4π/3}
[tex]Find all solutions to the equation sin2x+sinx-2cosx-1=0 in the interval [0, 2pi)[/tex]

A quadratic equation is one of the form ax2 + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0. So I think x = 4 but not 100%

Sorry if I'm wrong but I hope this helps

X=1, if thats what you want.

1/2

Step-by-step explanation:

EDG 2020

positive 3/2

Step-by-step explanation:

see explanation

Step-by-step explanation:

Given

2x² + x - 1 = 2 ( subtract 2 from both sides )

2x² + x - 3 = 0

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term

product = 2 × - 3 = - 6 and sum = + 1

The factors are - 2 and + 3

Use these factors to split the x- term

2x² - 2x + 3x - 3 = 0 ( factor the first/second and third/fourth terms )

2x(x - 1) + 3(x - 1) = 0 ← factor out (x - 1) from each term

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

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = - [tex]\frac{3}{2}[/tex]

The answer would be C. x=-3 or x=5.

You can find this answer using the zero product property and the quadratic formula. Hope this helps :)

Hello : 
2x²-x-15=x(x+1)
2x²-x-15-x(x+1) =0
2x²-x-15-x²-x =0
x²-2x-15 = 0
(x²-2x+1)-1-15 = 0
(x-1)² -4² =0 
Use the zero product property  and identity :a²-b²=(a-b)(a+b)
(x-1-4)(x-1+4) = 0
(x-5)(x+3) =0
x-5=0 or x+3 =0
x=5 or x= -3 

The answer is C There is explaining in the picture
[tex]What is the solution of this equation? 2x +x-11+3-7x=15 a.x=7/4 b.x=-7/4 c.x=-23/4 dx=23/4[/tex]

The value of X equals to 1