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Which value of x does not satisfy the equation sin^2x + sin x = 0

User Badunk
by
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2 Answers

4 votes

Final answer:

The values of x that do not satisfy the equation sin^2x + sin x = 0 are x = π, 2π, 3π, ..., -3π/2, -5π/2, ...

Step-by-step explanation:

The equation sin^2x + sin x = 0 is a quadratic equation in terms of sin x. Let's solve it:

  1. First, let's factor out sin x:
  2. sin x(sin x + 1) = 0
  3. Now, we have two cases: sin x = 0 or sin x + 1 = 0
  4. For the first case, sin x = 0, we have x = 0, π, 2π, 3π, ...
  5. For the second case, sin x + 1 = 0, we have sin x = -1, which gives x = -π/2, -3π/2, -5π/2, ...
  6. So the values of x that do not satisfy the equation sin^2x + sin x = 0 are x = π, 2π, 3π, ..., -3π/2, -5π/2, ...

User Tsilavina
by
7.9k points
3 votes

Answer:


x \\eq 0 \pm \pi \cdot i, \forall i \in \mathbb{N}_(O) and
x \\eq (3\pi)/(2)\pm 2\pi \cdot j, \forall j \in \mathbb{N}_(O).

Step-by-step explanation:

The equation can be simplified with the help of trigonometric identities:


\sin x \cdot (\sin x + 1) = 0

Which means that equation is equal to zero if any component of the product is zero. The solutions of the expression are:

a)
\sin x


x = 0 \pm \pi\cdot i,\forall i \in \mathbb{N}_(O)

b)
\sin x + 1


\sin x + 1 = 0


\sin x = -1


x = (3\pi)/(2) \pm 2\pi\cdot i,\forall i \in \mathbb{N}_(O)

Any value different of both subsets do not satisfy the equation described above.

User Khauri
by
7.3k points

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