If an acute angle $\theta$ satisfies the trigonometric identity $2{\sin }^2\theta -\cos \theta -1=0$, determine the value of $\tan \theta$.

The correct answer is $\sqrt{3}$.

Step-by-step explanation:

We are given the trigonometric identity for an acute angle $\theta$:

$2{\sin }^2\theta -\cos \theta -1=0$

Recall the fundamental Pythagorean trigonometric identity:

${\sin }^2\theta =1-{\cos }^2\theta$

Substitute this identity into the given equation to express it entirely in terms of $\cos \theta$:

$2(1-{\cos }^2\theta )-\cos \mathrm{\theta <mo>-</mo><mn>1</mn><mo>=</mo><mn>0</mn>}$

Expand and simplify the equation:

$2-2{\cos }^2\theta -\cos \mathrm{\theta <mo>-</mo><mn>1</mn><mo>=</mo><mn>0</mn>}$

$-2{\cos }^2\theta -\cos \theta +1=0$

Multiply the entire equation by -1 to make the leading coefficient positive:

$2{\cos }^2\theta +\cos \mathrm{\theta <mo>-</mo><mn>1</mn><mo>=</mo><mn>0</mn>}$

Factor the quadratic expression:

$(2\cos \theta -1)(\cos \theta +1)=0$

Set each factor equal to zero to find the roots:

1. $2\cos \theta -1=0\Rightarrow \cos \theta =\frac{1}{2}$

2. $\cos \theta +1=0\Rightarrow \cos \theta =-1$

Since $\theta$ is an acute angle ($0°<\theta <90°$), the cosine of $\theta$ must be positive. Therefore, we reject $\cos \theta =-1$ and accept:

$\cos \theta =\frac{1}{2}$

For an acute angle, $\cos \theta =\frac{1}{2}$ gives:

$\theta =60°$

Now, calculate the value of $\tan \theta$:

$\tan \theta =\tan (60°)=\sqrt{3}$

Thus, the value of $\tan \theta$ is $\sqrt{3}$.