Question

1. $sin^{2}\theta + cos\theta = 2$ (hint: use the pythagorean identity $sin^{2}\theta + cos^{2}\theta = 1$ to replace $sin^{2}\theta$ in the given equation). (4 points)

Explanation:

Step1: Substitute $\sin^2\theta$

$\sin^2\theta = 1-\cos^2\theta$, so substitute into equation:
$1-\cos^2\theta + \cos\theta = 2$

Step2: Rearrange to quadratic form

Bring all terms to left side:
$-\cos^2\theta + \cos\theta + 1 - 2 = 0$
$\cos^2\theta - \cos\theta + 1 = 0$

Step3: Solve quadratic equation

Let $x=\cos\theta$, equation becomes $x^2 - x + 1 = 0$.
Discriminant: $\Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3$

Step4: Analyze discriminant

Since $\Delta < 0$, no real $x$ exists. As $\cos\theta$ only takes real values between $-1$ and $1$, there is no real solution for $\theta$.

Answer:

There is no real solution for $\theta$.