Solve each equation for exact solutions over the interval [0,360)where appropriate. Round approximate solutions to the nearest tenth degree. Sin^2Theta=Cos^2Theta+1

My Work: Using double angle Identity I subtract 1 to other side therefore:
1-Sin^2theta=Cos^2theta= Cos2theta=Cos^2theta.

This is where I am stuck.For this does not conclude an answer...

since sin^2θ is never more than 1, you need cos^2θ = 0

so, θ = π/2 or 3π/2

or, solving algebraically,

sin^2θ = cos^2θ + 1
cos^2θ - sin^2θ = -1
cos(2θ) = -1
θ = π/2 or 3π/2

2sintheta-1=cscthet

To solve the equation sin^2(theta) = cos^2(theta) + 1, we can substitute the double angle identity for cosine: cos(2theta) = 1 - 2sin^2(theta).

So the equation becomes: sin^2(theta) = 1 - 2sin^2(theta) + 1.

Combining like terms, we have: 3sin^2(theta) = 2.

Divide both sides by 3 to isolate sin^2(theta): sin^2(theta) = 2/3.

Taking the square root of both sides, we get: sin(theta) = ±sqrt(2/3).

To find the exact solutions over the interval [0, 360), we need to consider the unit circle and identify when the sine function equals ±sqrt(2/3).

The exact solutions for sin(theta) = sqrt(2/3) are:

theta = 30 degrees, 150 degrees (in the first and second quadrants, where sin is positive).

The exact solutions for sin(theta) = -sqrt(2/3) are:

theta = 210 degrees, 330 degrees (in the third and fourth quadrants, where sin is negative).

So the exact solutions over the interval [0, 360) are: 30 degrees, 150 degrees, 210 degrees, and 330 degrees.