2 sinsqarepi/6=cosecsqare7pi/6 cossqare pie/3=3/2

To solve these trigonometric equations, we can use the identities and properties of trigonometric functions.

Let's start with the first equation:
2sin^2(π/6) = csc^2(7π/6)

We know that csc(θ) is equal to 1/sin(θ).

So, on the right side of the equation, we can rewrite csc^2(7π/6) as (1/sin(7π/6))^2.

Now let's solve the equation step by step:

1. First, let's find the value of sin(π/6).
The sine function of π/6 is equal to 1/2, so sin(π/6) = 1/2.

2. Next, let's find the value of sin(7π/6).
To find the sine of an angle greater than π/2, we can use the symmetry property of the sine function.
Since sin(π - θ) = sin(θ), we have sin(7π/6) = sin(π - π/6) = sin(5π/6).

Now, let's find the value of sin(5π/6):
The sine function of 5π/6 is 1/2, so sin(5π/6) = 1/2.

3. Now we have the values of sin(π/6) and sin(7π/6).
Let's substitute them into the original equation:
2sin^2(π/6) = (1/sin(7π/6))^2
2(1/2)^2 = (1/(1/2))^2
2(1/4) = (2)^2
1/2 = 4

The equation 1/2 = 4 is not true, so there is no solution to the equation 2sin^2(π/6) = csc^2(7π/6).

Moving on to the second equation:
cos^2(π/3) = 3/2

To solve this equation, we will use the identity cos^2(θ) + sin^2(θ) = 1.
By rearranging this identity, we get cos^2(θ) = 1 - sin^2(θ).

1. Let's find the value of sin(π/3).
The sine function of π/3 is √3/2, so sin(π/3) = √3/2.

2. Now, substitute the value of sin(π/3) into the identity we mentioned earlier:
cos^2(π/3) = 1 - sin^2(π/3)
cos^2(π/3) = 1 - (√3/2)^2
cos^2(π/3) = 1 - 3/4
cos^2(π/3) = 1/4

So, cos^2(π/3) is equal to 1/4, not 3/2.

In conclusion, the equation cos^2(π/3) = 3/2 does not have a true solution.