determine the solutions for tanxcos^2x-tanx=0 in the interval XE[-2pi,2pi]
Ok, please describe the steps to me that you used to solve this! thank you
tanx(cos^2 x - 1) = 0
tanx = 0 or cosx = ± 1
looking at tangents curves from -2pi to +2pi
x = -2pi , -pi, 0, pi, 2pi
looking at a cos curve where it is ±1
is -2pi, -pi 0, pi, and 2pi
so both parts had the same solutions,
x = 0, ±pi, ±2pi
Well, the first step I used was taking a big sip of coffee to wake up my math brain. Then, I took a deep breath and tackled the problem.
To solve the equation tan(x)cos^2(x) - tan(x) = 0, we can start by factoring out the common term, which is tan(x):
tan(x)(cos^2(x) - 1) = 0
Next, we can simplify the equation to:
tan(x)(sin^2(x)) = 0
Now, we have two factors: tan(x) = 0 and sin^2(x) = 0. Let's solve each factor separately.
For tan(x) = 0, we know that x = 0 is a solution. But since we're considering the interval [-2π, 2π], we also need to look for solutions where tan(x) is zero within this range.
The solutions for tan(x) = 0 in the given interval are x = -π, 0, and π.
Now, let's look at the second factor, sin^2(x) = 0. Since sin^2(x) is zero when sin(x) is zero, we need to find when sin(x) is zero in the given interval.
The solutions for sin(x) = 0 in the interval [-2π, 2π] are x = -2π, -π, 0, π, and 2π.
Combining all the solutions for both factors, we have x = -2π, -π, 0, π, and 2π.
And voila! We've found the solutions for the equation tan(x)cos^2(x) - tan(x) = 0 in the interval [-2π, 2π]. Just be careful not to trip over any tangents along the way!
To solve the equation tan(x)cos^2(x) - tan(x) = 0 in the interval x ∈ [-2π, 2π], follow these steps:
Step 1: Factor out tan(x) from the equation:
tan(x)(cos^2(x) - 1) = 0
Step 2: Observe that cos^2(x) - 1 = sin^2(x) using the identity cos^2(x) + sin^2(x) = 1.
Step 3: Rewrite the equation as:
tan(x)sin^2(x) = 0
Step 4: Set each factor to zero and solve for x:
tan(x) = 0
x = 0, π, -π
sin(x) = 0
x = 0, π
Step 5: Combine the solutions obtained:
x = 0, π, -π
Step 6: Check if these solutions lie within the given interval [-2π, 2π].
0, π, and -π all fall within the interval [-2π, 2π]. Therefore, the solutions to the equation within the given interval are:
x = 0, π, -π.
To solve the equation tan(x)cos^2(x) - tan(x) = 0 in the interval x ∈ [-2π, 2π], follow these steps:
Step 1: Factor out the common term tan(x) from the equation:
tan(x)(cos^2(x) - 1) = 0
Step 2: Apply the zero product property: if a product of factors is equal to zero, then at least one of the factors must be zero. So we set each factor equal to zero and solve for x.
tan(x) = 0 or cos^2(x) - 1 = 0
Step 3: Solve the first equation tan(x) = 0:
From the unit circle, we know that tan(x) = 0 when x is an integer multiple of π. In the given interval [-2π, 2π], the solutions are x = 0, π, and -π.
Step 4: Solve the second equation cos^2(x) - 1 = 0:
Rearrange the equation as cos^2(x) = 1.
Step 5: Take the square root of both sides of the equation:
cos(x) = ±1
Step 6: Use the unit circle to find the values of x where cos(x) = ±1:
From the unit circle, cos(x) = 1 when x is an even multiple of π.
And cos(x) = -1 when x is an odd multiple of π.
In the given interval [-2π, 2π], the solutions for cos(x) = 1 are x = 0, 2π, -2π.
The solutions for cos(x) = -1 are x = π, -π.
So, combining all the solutions:
x = 0, π, -π, 2π, -2π.
These are the solutions for the equation tan(x)cos^2(x) - tan(x) = 0 in the interval x ∈ [-2π, 2π].