Solve using the quadratic formula:

\(\displaystyle \cos x = \frac{21 \pm \sqrt{441 - 4(4)(5)}}{2(4) } = \frac{ 21 \pm \sqrt{361}}{8} = \frac{21 \pm 19}{8}\)

\(\displaystyle \cos x = 5, \frac{1}{4}\)

5 is outside the range for cosine, so the only solution that works is \(\displaystyle \cos x = \frac{1}{4}\):

\(\displaystyle x = \cos ^{-1} (\frac{1}{4}) \approx 75.52 ^o\) according to a calculator

The other angle with a cosine of \(\displaystyle \frac{1}{4}\) is \(\displaystyle 360^o - 75.52^o = 284.48^o\)

Use the quadratic formula:

\(\displaystyle \cos x = \frac{ -8 \pm \sqrt{64 - 4(3)(4)}}{2(3)} = \frac{-8 \pm \sqrt{16}}{6 } = \frac{-8\pm4}{6}\)

\(\displaystyle \cos x = -2 \enspace or - \frac{2}{3}\)

-2 is outside the range of cosine, so the answer has to come from \(\displaystyle - \frac{2}{3}\):

\(\displaystyle \cos x = - \frac{2}{3}\)

\(\displaystyle x = \cos^{-1} (-\frac{2}{3} ) \approx 131.81^o\) according to a calculator

The other angle with a cosine of \(\displaystyle -\frac{2}{3}\) is \(\displaystyle 360^o - 131.81^o = 228.19^o\)

First of all, we can use the Pythagorean identity \(\displaystyle sin^2x+cos^2x=1\) to rewrite the given equation in terms of \(\displaystyle sinx\).

\(\displaystyle 2cos^2x+sinx=1\)

\(\displaystyle \Rightarrow 2(1-sin^2x)+sinx=1\)

\(\displaystyle \Rightarrow 2-2sin^2x+sinx=1\)

\(\displaystyle \Rightarrow -2sin^2x+sinx+1=0\)

\(\displaystyle \Rightarrow 2sin^2x-sinx-1=0\)

This is a quadratic equation in terms of \(\displaystyle sinx\); hence, we can use the quadratic formula to solve this equation for \(\displaystyle sinx\).

\(\displaystyle sinx=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

where \(\displaystyle a=2, b=-1, c=-1\).

\(\displaystyle \Rightarrow sinx=\frac{1\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}\)

\(\displaystyle \Rightarrow sinx=\frac{1\pm3}{4}\)

\(\displaystyle \Rightarrow sinx=-\frac{1}{2}, sinx=1\).

Now, \(\displaystyle sinx=1\) when \(\displaystyle x=\frac{\pi}{2}\), and \(\displaystyle sinx=-\frac{1}{2}\) when \(\displaystyle x=\frac{7\pi}{6}\) or \(\displaystyle \frac{11\pi}{6}\).

Hence, the solutions to the original equation \(\displaystyle 2cos^2x+sinx=1\) are

\(\displaystyle x=\) \(\displaystyle \frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}\)

We start by rewriting the \(\displaystyle \cos^2x\) term on the right hand side in terms of \(\displaystyle \sin^2x\).

\(\displaystyle \cos^2x=1-\sin^2x\)

\(\displaystyle 1-\sin x=1-\sin^2x-\frac{1}{4}\)

We then move everything to the left hand side of the equation and cancel.

\(\displaystyle 1-\sin x-1+\sin^2x+\frac{1}{4}=0\)

\(\displaystyle \sin^2x-\sin x+\frac{1}{4}=0\)

Apply the quadratic formula:

\(\displaystyle \begin{align*} \sin x&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(\frac{1}{4})}}{2(1)}\\ &=\frac{1\pm\sqrt{1-1}}{2}\\ &=\frac{1}{2} \end{align*}\)

So \(\displaystyle \sin x=\frac{1}{2}\). Using the unit circle, the two values of \(\displaystyle x\) that yield this are \(\displaystyle x=\frac{\pi}{6}\) and \(\displaystyle x=\frac{5\pi}{6}\).

Since the triangle in question is a right triangle we can use the Pythagorean Theorem. First, we must determine which sides we are given. Since the function we are given is sine, we know that we are given the opposite side and the hypotenuse. Therefore, setting up the equation:

\(\displaystyle a^{2}+b^{2}=c^{2}\)

Where, \(\displaystyle a\) and \(\displaystyle c\) are given.

\(\displaystyle 3^{2} + b^{2} = 5^{2}\)

Solving the above equation:
\(\displaystyle b = \sqrt{16} = 4\)

We toss out the negative solution since the length of a side must be positive.

Since the triangle in question is a right triangle we can use the Pythagorean Theorem. First, we must determine which sides we are give. Since the function we are given is tangent, we know that we are given the opposite and adjacent sides. Therefore, setting up the equation:

\(\displaystyle a^{2}+b^{2}=c^{2}\)

Where, \(\displaystyle a\) and \(\displaystyle b\) are given.

\(\displaystyle 17^{2}+18^{2} = c^{2}\)

Solving the above equation:

\(\displaystyle c = \sqrt{613} = 24.76\)

We toss out the negative solution since the length of a side must be positive.

Since the triangle in question is a right triangle we can use the Pythagorean Theorem. First, we must determine which sides we are given. Since the function we are given is cosine, we know that we are given the adjacent side and hypotenuse. Therefore, setting up the equation:

\(\displaystyle a^{2}+b^{2}=c^{2}\)

Where, \(\displaystyle a\) and \(\displaystyle c\) are given.

\(\displaystyle 104^{2} + b^{2} = 109^{2}\)

Solving the above equation:

\(\displaystyle b=\sqrt{1,065}=32.63\)

We toss out the negative solution since the length of a side must be positive.

We are given the opposite side, with respect to the angle, along with the angle. Therefore, we utilize the sine function to determine the length of the hypotenuse:

\(\displaystyle \sin\theta = \frac{opposite}{hypotenuse}\)

Substituting the given values:

\(\displaystyle \sin 17^{\circ} = \frac{6}{h}\)

Cross multiplying:

\(\displaystyle h\sin 17^{\circ}=6\)

Solving for \(\displaystyle h\):

\(\displaystyle h = \frac{6}{\sin 17^{\circ}} = 20.52\)

We are given two sides of the right triangle, namely the hypotenuse and the opposite side of the angle. Therefore, we simply use the sine function to determine the angle:
\(\displaystyle \sin\theta = \frac{opposite}{hypotenuse} =\frac{7}{13}\)

In order to isolate the angle we must apply the inverse sine function to both sides:

\(\displaystyle \theta = sin^{-1}\frac{7}{13} = 32.57^{\circ} \approx 33^{\circ}\)

We are given two sides of the right triangle, namely the hypotenuse and the adjacent side of the angle. Therefore, we simply use the cosine function to determine the angle:

\(\displaystyle \cos\theta = \frac{adjacent}{hypotenuse}=\frac{35}{70}\)

In order to isolate the angle we must apply the inverse cosine function to both sides:

\(\displaystyle \theta = \cos^{-1}\frac{35}{70}=\cos^{-1}\frac{1}{2}=60^{\circ}\)