Find all solutions to the given equation in the interval [0,2π). Give the exact solution, including "pi" for π. For any unused answer boxes, enter DNE in all capital letters.
2tanxsinx+2tanx=0
explanation please...thank you much
really?
2tanx sinx + 2tanx = 0
2tanx (sinx + 1) = 0
tanx = 0
sinx = -1
You should be able to take it from there.
and when posting a question, don't bother with things like
Give the exact solution, including "pi" for π. For any unused answer boxes, enter DNE in all capital letters.
Instructions for input format are for you -- they have nothing to do with solving the problem.
sorry we are not as smart as you Oobleck
but thank you
To find the solutions to the given equation, we'll start by factoring out the common term, 2tan(x), from both terms of the equation:
2tan(x)sin(x) + 2tan(x) = 0
Now, we can factor out 2tan(x) from the left side of the equation:
2tan(x)(sin(x) + 1) = 0
For the product of two factors to equal zero, one or both of the factors must be zero. Therefore, we set each factor equal to zero and solve for x.
Setting 2tan(x) equal to zero:
2tan(x) = 0
Dividing both sides of the equation by 2 gives:
tan(x) = 0
To determine the exact solutions, we can look at the unit circle or the values of tangent on the interval [0,2π). The tangent function equals zero at x = 0 and x = π since these are the angles where the sine function is zero.
Next, setting (sin(x) + 1) equal to zero:
sin(x) + 1 = 0
Subtracting 1 from both sides of the equation:
sin(x) = -1
To determine the exact solutions, we can look at the unit circle or the values of sine on the interval [0,2π). The sine function equals -1 at x = 3π/2.
Therefore, the solutions to the given equation in the interval [0,2π) are:
x = 0, π, and 3π/2.
Please let me know if I can help you with anything else.