2sec^2x -3tanx =5 for [0,2pi)
Please solve for the values of x for me.
I am sure bobpursly had already answered this for you, but I could not find the post. I recall he took you as far as the quadratic. Why did you not finish it ??
2sec^2x -3tanx =5 for [0,2pi)
make use of your identities
sec^2 A = 1 + tan^2 A
2(1 + tan^2 x) - 3tanx - 5 = 0
2 + 2tan^2 x - 3tanx - 5 = 0
2tan^2 x - 3tanx - 3 = 0
by the formula:
tanx = (3 ± √33)/4
= appr 2.18614... or -.68614...
using tanx = 2.18614...
x is in quads I or III
x = 1.1418 or x = π + 1.1418 = 4.2834
using tanx = -.68614..
x is in quads II or IV
x = π - .60136... = 2.5102
x - 2π - .60136 = 5.6818 , all 4 answers rounded to 4 decimal at the final stage, not before.
in the line near the end:
x = π - .60136... = 2.5102
it should be
x = π - .60136... = 2.5402>/b>
To solve the equation 2sec^2x -3tanx = 5 for the values of x in the interval [0, 2π), follow these steps:
Step 1: Rewrite the equation in terms of sine and cosine using the identity sec^2x = 1 + tan^2x.
2(1 + tan^2x) - 3tanx = 5
Step 2: Distribute the 2 to both terms inside the parentheses.
2 + 2tan^2x - 3tanx = 5
Step 3: Rearrange the equation to form a quadratic equation.
2tan^2x - 3tanx - 3 = 0
Step 4: Factor the quadratic equation.
(2tanx + 1)(tanx - 3) = 0
Step 5: Set each factor equal to zero and solve for x separately.
2tanx + 1 = 0 OR tanx - 3 = 0
Step 6: Solve the first equation.
2tanx = -1
tanx = -1/2
To find the values of x within the interval [0, 2π) where tanx = -1/2, you can use the unit circle or a calculator. The solutions are approximately x ≈ 7π/6 and x ≈ 11π/6.
Step 7: Solve the second equation.
tanx = 3
To find the values of x within the interval [0, 2π) where tanx = 3, you can again use the unit circle or a calculator. The solutions are approximately x ≈ π/4 and x ≈ 5π/4.
Therefore, the values of x that satisfy the equation 2sec^2x -3tanx = 5 in the interval [0, 2π) are x ≈ 7π/6, x ≈ 11π/6, x ≈ π/4, and x ≈ 5π/4.
To solve the equation 2sec^2x - 3tanx = 5 for the values of x in the interval [0, 2π), we will go through the following steps:
Step 1: Simplify the equation using trigonometric identities.
Step 2: Solve the resulting quadratic equation.
Step 3: Check the solutions in the given interval.
Let's start with step 1:
Step 1: Simplify the equation using trigonometric identities.
Recall the following identities:
- sec^2x = 1 + tan^2x
- tanx = sinx/cosx
Using these identities, we can rewrite the equation as follows:
2(1 + tan^2x) - 3(tanx) = 5
Simplifying further:
2 + 2tan^2x - 3tanx = 5
Rearranging the equation:
2tan^2x - 3tanx - 3 = 0
Now, we move to step 2:
Step 2: Solve the resulting quadratic equation.
To solve the quadratic equation, let's substitute tanx with a variable, u. Thus, the equation becomes:
2u^2 - 3u - 3 = 0
Now we can apply the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
u = (-(-3) ± √((-3)^2 - 4(2)(-3))) / (2(2))
Simplifying:
u = (3 ± √(9 + 24)) / 4
u = (3 ± √33) / 4
So, we have two possible values for u:
u = (3 + √33) / 4 (Solution 1)
u = (3 - √33) / 4 (Solution 2)
Now let's move to step 3:
Step 3: Check the solutions in the given interval.
To check if the solutions are within the given interval [0, 2π), substitute u back into the original equation:
For Solution 1:
tanx = (3 + √33) / 4
x = arctan((3 + √33) / 4)
For Solution 2:
tanx = (3 - √33) / 4
x = arctan((3 - √33) / 4)
Evaluate these inverse tangent values using a calculator or computer program:
x ≈ 1.059, 3.032
Therefore, the values of x that satisfy the equation 2sec^2x - 3tanx = 5 in the interval [0, 2π) are approximately 1.059 and 3.032.