Use the quadratic formula to solve for \(\displaystyle \cos (3 \theta )\):

\(\displaystyle \cos (3 \theta ) = \frac{ -5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{ -5 \pm \sqrt{49}}{4} = \frac{ -5 \pm 7 }{4}\)

One possible solution is: 

\(\displaystyle \cos (3 \theta ) = \frac{- 5 - 7 }{4} = \frac{-12}{4} = -3\) this is outside of the possible range for cosine

The other solution is:

\(\displaystyle \cos (3 \theta ) = \frac{ -5 + 7 }{4} = \frac{ 2}{4} =\frac{ 1}{2}\) 

\(\displaystyle 3 \theta = \cos ^ {-1 } (\frac{1}{2} ) = \frac{ \pi }{3} , \frac{ 5 \pi }{3}\) divide by 3

\(\displaystyle \theta = \frac{ \pi }{ 9 } , \frac{ 5 \pi }{ 9 }\)

Solve using the quadratic formula:

\(\displaystyle \sin ^ 2 \theta = \frac{ -9 \pm \sqrt{ 81 - 4(4)(-9)}}{2(4)} = \frac{ -9 \pm \sqrt{225}}{8 } = \frac{ -9 \pm 15}{8}\)

One possible answer is:

\(\displaystyle \sin ^ 2 \theta = \frac{ - 9 + 15 }{ 8 } = \frac{ 6 }{ 8 } = \frac{ 3}{4}\) take the square root

\(\displaystyle \sin \theta = \pm \sqrt{ \frac{ 3}{4} } = \pm \frac{ \sqrt3}{2}\)

\(\displaystyle \theta = \sin ^ {-1} ( \pm \frac{ \sqrt3}{2} )\)

\(\displaystyle \theta = \frac{ \pi }{3} , \frac{2 \pi }{3} , \frac{ 4 \pi }{3} , \frac{ 5 \pi }{3}\)

The other would be:

\(\displaystyle \sin ^ 2 \theta = \frac{ -9 - 15 }{8 } = \frac{ - 24}{8 } = -3\) this is outside of the range for sine

Subtracting 5 from both sides gives the quadratic equation \(\displaystyle 0 = 2 \cos ^2 \theta - 5 \cos \theta - 3\)

Using the quadratic formula gives:

\(\displaystyle \cos \theta = \frac{ 5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{ 5 \pm 7}{4} = \frac{ -1}{2} , 3\)

The cosine cannot be 3 because that's greater than 1.

\(\displaystyle \cos \theta = - \frac{1}{2}\)

\(\displaystyle \theta = \cos ^{-1} (-\frac{1}{2}) = \frac{2 \pi }{3} , \frac{ 4 \pi }{3}\)

Using the quadratic formula gives:

\(\displaystyle \sin \theta = \frac{ -1 \pm \sqrt{ 1 - 4(10)(-3)}}{2(10)} = \frac{-1 \pm 11}{20 } = \frac{1}{2} , -\frac{3}{5}\)

\(\displaystyle \theta = \sin ^{-1} (\frac{ 1}{2} ) = 30^o, 150^o\) or

\(\displaystyle \theta = \sin ^{-1} (-\frac{3}{5}) \approx 323.13^o, 143.13^o\)

Solve using the quadratic formula:

\(\displaystyle \sin \theta = \frac{ 3 \pm \sqrt{9 - 4(2)}}{4(1)} = \frac{ 3 \pm \sqrt1}{4} = \frac{1}{2} \enspace or \enspace 1\)

\(\displaystyle \sin \theta = \frac{1}{2}\)

\(\displaystyle \theta = \sin ^{-1} (\frac{1}{2} ) = \frac{\pi }{6} , \frac{ 5 \pi }{6}\)

\(\displaystyle \sin \theta = 1\)

\(\displaystyle \theta = \sin ^{-1}(1)\)

\(\displaystyle \theta = \frac{\pi }{2}\)

To solve, use the quadratic formula:

\(\displaystyle \sin \theta = \frac{ -2 \pm \sqrt{4 + 20 }}{2 } = \frac{-2 \pm \sqrt{24}}{2} = - \pm \sqrt{6}\)

Both \(\displaystyle -1 - \sqrt{6}\) and \(\displaystyle -1+\sqrt{6}\) are outside of the range of the sine function, \(\displaystyle -1 \leq \sin x \leq 1\) so there is no solution.

Solve using the quadratic formula:

\(\displaystyle \cos x = \frac{-1 \pm \sqrt{1 + 16}}{4 } = \frac{-1 \pm \sqrt{17}}{4}\)

 \(\displaystyle \frac{ -1 - \sqrt{17}}{4} \approx -1.28\), outside the range for cosine.

\(\displaystyle \frac{-1 + \sqrt{17}}{4} \approx 0.78\)

\(\displaystyle x = \cos ^{-1} (0.78 ) = 38.67^o\) according to a calculator.

The other angle with a cosine of 0.78 would be \(\displaystyle 360 ^o - 38.67^o = 321.33^o\).

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

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