Question

How do you solve $ \sin 2\theta - \cos \theta = 0 $ between $ 0 $ and $ 2\pi $ ?

Answer 1

Hint: First, we know that double angle identities are special cases of the sum identities. That is, when the two angles are equal, the sum identities are reduced to double angle identities.
The double sine formula is,
 $ \sin 2A = 2\sin A\cos A $
We will substitute this double angle formula in the given equation. Then, we will take sin common and keep each factor equal to $ 0 $ .
After we solve for $ \cos \theta $ and find the value of $ \theta $
And finally, we will find the value of $ \theta $ .
 $ \sin \theta $ will repeat after every pie, so we find $ 0 $ and $ 2\pi $ .

Complete step-by-step solution:
The given trigonometry is $ \sin 2\theta - \cos \theta = 0 $
We know that $ \sin 2A = 2\sin A\cos A $
Convert $ A $ into $ \theta $ , hence we get
$\Rightarrow$ $ \sin 2\theta = 2\sin \theta \cos \theta $
Putting this in the given equation
$\Rightarrow$ $ 2\sin \theta cos\theta - \cos \theta = 0 $
Now, we take the common term $ \cos \theta $
$\Rightarrow$ $ \cos \theta (2\sin \theta - 1) = 0 $
If any individual factor on the left side of the equation is equal to zero, the entire expression will be equal to zero, hence we get
$\Rightarrow$ $ \cos (\theta ) = 0 $
$\Rightarrow$ $ 2\sin (\theta ) - 1 = 0 $
Set the first factor equal to zero and solve
$\Rightarrow$ $ \cos (\theta ) = 0 $
Take the inverse cosine of both sides of the equation to extract $ \theta $ from inside the cosine.
$\Rightarrow$ $ \theta = \arccos (0) $
The exact value of $ \arccos (0) $ is $ \dfrac{\pi }{2} $
$\Rightarrow$ $ \theta = \dfrac{\pi }{2} $
The cosine function is positive in the first and fourth quadrant. To find the second solution, subtract the reference angle from $ 2\pi $ to find the solution in the fourth quadrant.
$\Rightarrow$ $ \theta = 2\pi - \dfrac{\pi }{2} $
Simplify $ 2\pi - \dfrac{\pi }{2} $
To write $ \dfrac{{2\pi }}{1} $ as a fraction with a common denominator, multiply by $ \dfrac{2}{2} $ .
$\Rightarrow$ $ \theta = \dfrac{{2\pi }}{1} \times \dfrac{2}{2} - \dfrac{\pi }{2} $
Combine
$\Rightarrow$ $ \theta = \dfrac{{2\pi \cdot 2}}{{1 \cdot 2}} - \dfrac{\pi }{2} $
Multiply $ 2 $ by $ 1 $
$\Rightarrow$ $ \theta = \dfrac{{2\pi \cdot 2}}{2} - \dfrac{\pi }{2} $
Combine the numerators over the common denominator
$\Rightarrow$ $ \theta = \dfrac{{2\pi \cdot 2 - \pi }}{2} $
Simplify the numerator
Multiply $ 2 $ by $ 2 $
$\Rightarrow$ $ \theta = \dfrac{{4\pi - \pi }}{2} $
Subtract $ \pi $ from $ 4\pi $
$\Rightarrow$ $ \theta = \dfrac{{3\pi }}{2} $
The period of the function can be calculated using $ \dfrac{{2\pi }}{{\left| b \right|}} $
Replace $ b $ with $ 1 $ in the formula for a period
$\Rightarrow$ $ \dfrac{{2\pi }}{{\left| 1 \right|}} $
The absolute value is the distance between a number and zero. The distance between $ 0 $ and $ 1 $ is $ 1 $
$\Rightarrow$ $ \dfrac{{2\pi }}{1} $
Divide $ 2\pi $ by $ 1 $
$\Rightarrow$ $ 2\pi $
The period of the $ \cos (\theta ) $ function is $ 2\pi $ so values will repeat every $ 2\pi $ radian in both directions.
$\Rightarrow$ $ \theta = \dfrac{\pi }{2} + 2\pi n,\dfrac{{3\pi }}{2} + 2\pi n, $ for any integer $ n $
Set the next factor equal to zero and solve
$\Rightarrow$ $ 2\sin (\theta ) - 1 = 0 $
Add $ 1 $ to both sides of the equation
$\Rightarrow$ $ 2\sin (\theta ) = 1 $
Divide each term by $ 2 $ and simplify
$\Rightarrow$ $ \sin (\theta ) = \dfrac{1}{2} $
Take the inverse sine of both sides of the equation to extract $ \theta $ from inside the sine.
$\Rightarrow$ $ \theta = \arcsin \left( {\dfrac{1}{2}} \right) $
The exact value of $ \arcsin \left( {\dfrac{1}{2}} \right) $ is $ \dfrac{\pi }{6} $
$\Rightarrow$ $ \theta = \dfrac{\pi }{6} $
The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $ \pi $ to find the solution in the second quadrant.
$\Rightarrow$ $ \theta = \pi - \dfrac{\pi }{6} $
Simplify
To write $ \dfrac{\pi }{1} $ as a fraction with a common denominator, multiply by $ \dfrac{6}{6} $ .
$\Rightarrow$ $ \theta = \dfrac{\pi }{1} \cdot \dfrac{6}{6} - \dfrac{\pi }{6} $
Combine the numerator over the denominator
$\Rightarrow$ $ \theta = \dfrac{{\pi \cdot 6 - \pi }}{6} $
Simplify the numerator
$\Rightarrow$ $ \theta = \dfrac{{5\pi }}{6} $
The period of the function can be calculated
$\Rightarrow$ $ 2\pi $
The period of the $ \sin (\theta ) $ function is $ 2\pi $ so values will repeat every $ 2\pi $ radian in both directions.
$\Rightarrow$ $ \theta = \dfrac{\pi }{6} + 2\pi n,\dfrac{{5\pi }}{6} + 2\pi n, $ for any integer $ n $
The final solution is all the values that make
 $ \cos (\theta )(2\sin (\theta ) - 1) = 0 $ true.
$\Rightarrow$ $ \theta = \dfrac{\pi }{2} + 2\pi n,\dfrac{{3\pi }}{2} + 2\pi n,\dfrac{\pi }{6} + 2\pi n,\dfrac{{5\pi }}{6} + 2\pi n, $ for any integer $ n $
Consolidate $ \dfrac{\pi }{2} + 2\pi n $ and $ \dfrac{{3\pi }}{2} + 2\pi n $ to $ \dfrac{\pi }{2} + \pi n $

Therefore $ \theta = \dfrac{\pi }{2} + \pi n,\dfrac{\pi }{6} + 2\pi n,\dfrac{{5\pi }}{6} + 2\pi n $ for any integer $ n $

Note: Application of $ \sin 2A = 2\sin A\cos A $ : When an object is projected with speed $ u $ at an angle $ \alpha $ to the horizontal over level, the horizontal distance (Range) it travels before striking the ground is given by the formula
 $ R = \dfrac{{{u^2}\sin 2\alpha }}{g} $
Clearly maximum of $ R $ is $ \dfrac{{{u^2}}}{g} $ , when $ \alpha = \dfrac{\pi }{4} $