Solve 6sin^2 x-5cosx-2=0 in the interval 0≤x≤2π
6sin^2 x-5cosx-2=0
6(1 - cos^2 x) - 5cosx - 2 = 0
6 - 6cos^2 x - 5cosx - 2 = 0
6cos^2 x + 5cosx -4 = 0
(2cosx - 1)(3cosx + 4) = 0
cosx = 1/2 or cosx = -4/3, but the latter is not possible since -1 ≤ cosx < +1
x = 60° or x = 300°
in radians: x = π/3 or 5π/3
How to get the x=60° or x=300°? I didn't catch that point.
Why did the chicken go to the seance? To communicate with its ghost ancestor and solve trigonometric equations, of course! Let's get cracking!
To solve the equation 6sin^2 x - 5cosx - 2 = 0, we can use some trigonometric wizardry and turn those sines and cosines into a single trig function.
First, let's rewrite sin^2 x in terms of cosx:
sin^2 x = 1 - cos^2 x.
Replacing sin^2 x in the original equation, we get:
6(1 - cos^2 x) - 5cosx - 2 = 0.
Expanding and rearranging terms, we have:
6 - 6cos^2 x - 5cosx - 2 = 0.
Now, let's simplify things further by combining like terms:
-6cos^2 x - 5cosx + 4 = 0.
To make our lives easier, let's substitute y for cosx:
-6y^2 - 5y + 4 = 0.
Now we have a nice quadratic equation in y! Let's solve it using our trusty friend, the quadratic formula:
y = (-b ± √(b^2 - 4ac))/(2a).
Plugging in the values a = -6, b = -5, and c = 4, we get:
y = (-(-5) ± √((-5)^2 - 4(-6)(4)))/(2(-6)).
Simplifying further, we have:
y = (5 ± √(25 + 96))/(-12).
Now, to find the values of cosx, we need to substitute y back in:
cosx = y.
Let's break out the calculator and calculate the two possible values of cosx:
cosx ≈ 0.826 and cosx ≈ -0.386.
Since we're looking for solutions in the interval 0 ≤ x ≤ 2π, we need to determine the corresponding values of x.
For cosx ≈ 0.826, we have x ≈ 0.586 radians or x ≈ 5.697 radians.
For cosx ≈ -0.386, we have x ≈ 2.013 radians or x ≈ 4.888 radians.
And there you have it, dear seeker of trigonometric solutions! The equation 6sin^2 x - 5cosx - 2 = 0 is solved, with solutions x ≈ 0.586, x ≈ 2.013, x ≈ 4.888, and x ≈ 5.697. Now, off to the seance with you!
To solve the equation 6sin^2(x) - 5cos(x) - 2 = 0 in the interval 0 ≤ x ≤ 2π, we can follow these steps:
Step 1: Rewrite the equation using the identity sin^2(x) + cos^2(x) = 1.
6(1 - cos^2(x)) - 5cos(x) - 2 = 0
Step 2: Distribute and simplify.
6 - 6cos^2(x) - 5cos(x) - 2 = 0
6cos^2(x) + 5cos(x) - 4 = 0
Step 3: Let u = cos(x) and solve the equation.
6u^2 + 5u - 4 = 0
Step 4: Solve the quadratic equation 6u^2 + 5u - 4 = 0.
This can be factored as:
(2u - 1)(3u + 4) = 0
Step 5: Set each factor equal to zero and solve for u.
2u - 1 = 0 ---> u = 1/2
3u + 4 = 0 ---> u = -4/3
Step 6: Use the inverse cosine function to find the values for x.
cos(x) = 1/2 ---> x = π/3, 5π/3
cos(x) = -4/3 (Not possible since the range of cos(x) is -1 to 1)
Step 7: Check the solutions in the original equation.
For x = π/3:
6sin^2(π/3) - 5cos(π/3) - 2 = 6(3/4) - 5(1/2) - 2 = 9/2 - 5/2 - 4/2 = 0 (True)
For x = 5π/3:
6sin^2(5π/3) - 5cos(5π/3) - 2 = 6(3/4) - 5(-1/2) - 2 = 9/2 + 5/2 - 4/2 = 5 (Not True)
Step 8: Final Answer
The solution to the equation 6sin^2(x) - 5cos(x) - 2 = 0, in the interval 0 ≤ x ≤ 2π, is x = π/3.
To solve the equation 6sin^2x - 5cosx - 2 = 0 in the interval 0≤x≤2π, we can use some trigonometric identities and properties.
First, let's rewrite the equation in terms of sinx only. We know that cos^2x + sin^2x = 1, so we can rewrite cosx as √(1 - sin^2x).
Our equation becomes:
6sin^2x - 5√(1 - sin^2x) - 2 = 0
Let's substitute a variable, say u, for sinx. So now our equation becomes:
6u^2 - 5√(1 - u^2) - 2 = 0
To simplify this equation further, let's square both sides:
(6u^2 - 2)^2 = (5√(1 - u^2))^2
36u^4 - 24u^2 + 4 = 25(1 - u^2)
Expanding both sides, we get:
36u^4 - 24u^2 + 4 = 25 - 25u^2
Rearranging terms, we have:
36u^4 - 24u^2 + 25u^2 - 4 = 25
36u^4 + u^2 - 29 = 0
Now, we can solve this quadratic equation by substituting a variable, say v, for u^2:
36v^2 + v - 29 = 0
To find the values of v, we can use the quadratic formula:
v = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 36, b = 1, and c = -29. Substituting these values into the quadratic formula, we get:
v = (-(1) ± √((1)^2 - 4(36)(-29))) / (2(36))
v = (-1 ± √(1 + 4176)) / 72
v = (-1 ± √4177) / 72
Now, we solve for u by taking the square root of v:
u = ± √((v + 1) / 36)
u = ± √(((√4177-1)/72 + 1) / 36)
u = ± √((√4177-1+72)/72*36)
Now, we substitute u = sinx back into the equation. Since sinx ranges from -1 to 1, we only consider the positive square root:
sinx = √((√4177-1+72)/72*36)
To find x, we can take the inverse sine (arcsin) of both sides:
x = arcsin(√((√4177-1+72)/72*36))
Now, we can plug this expression into a calculator to find the value of x in the interval 0≤x≤2π.