how do i solve this using trig identities?
2 csc^2x = 3 cot^2x-1
first change everything to sines and cosines
2/(sin^2 x) = 3(cos^2 x)/(sin^2 x) - 1
multiply by sin^2 x
2 = 3 cos^2 x - sin^2 x
but sin^2 x - 1 - cos^2 x
so
2 = 3 cos^2 x - 1 + cos^2 x
3 = 4cos^2 x
cos^2 x = 3/4
cos x = ±√3/2
Do you recognize the 1,√3,2 right-angled triangle? If not your calculuator should give you 30º.
if cos x = +√3/2 then x is in the first our fourth quadrants and
x = 30º or 330º
if cos x = -√3/2 then x is in the second and third quadrants and
x = 150º or 210º
(in radians the answers would be
pi/6, 11pi/6, 5pi/6,7pi/6)
To solve the equation using trigonometric identities, we need to manipulate the equation until we reach an identity or equation that we can solve.
Let's start by expressing cosecant squared (csc^2) and cotangent squared (cot^2) in terms of sine and cosine using the fundamental trigonometric identities:
csc^2x = 1/sin^2x
cot^2x = cos^2x/sin^2x
Now, let's substitute these expressions back into the original equation:
2(1/sin^2x) = 3(cos^2x/sin^2x) - 1
Next, let's simplify the equation by multiplying both sides by sin^2x to get rid of the denominators:
2 = 3cos^2x - sin^2x
Now, let's use the Pythagorean identity (sin^2x + cos^2x = 1) to rewrite sin^2x in terms of cos^2x:
2 = 3cos^2x - (1 - cos^2x)
Simplifying the equation further:
2 = 4cos^2x - 1
Rearranging the terms:
4cos^2x = 2 + 1
4cos^2x = 3
Dividing both sides by 4:
cos^2x = 3/4
To solve for cosx, we take the square root of both sides:
cosx = ±√(3/4)
Now, to find the values of x, we need to consider the range of the arccosine function. Since cosx can take values between -1 and 1, and cosx = ±√(3/4) is positive, we can take the positive square root.
cosx = √(3/4)
To find the values of x, we take the inverse cosine (arccos) of both sides:
x = arccos(√(3/4))
The exact values might depend on the specific range or interval you are considering, so you can use a calculator or trigonometric tables to find an approximate value for x.