Solving a Tangent Equation or Inequality

1. Isolate the tangent ratio.
   ||\begin{align}\tan(2x)-3&=-2\\\tan(2x)&=1\end{align}||

2. Determine the trigonometric angle.

3. Solve the equation.
   ||\begin{align}\tan(2x)&=1\\ &\Downarrow\\2x&=\dfrac{\pi}{4}\\[3pt]\boldsymbol{\color{#fa7921}{x_1}}&=\boldsymbol{\color{#fa7921}{\dfrac{\pi}{8}}}\end{align}||

4. Calculate the period of the tangent function.

The tangent function is periodic, so we need to calculate the period in order to determine all the solutions.||\begin{align}p&=\dfrac{\pi}{\vert b\vert}\\[3pt]&=\dfrac{\pi}{\vert2\vert}\\[3pt]&=\dfrac{\pi}{2}\end{align}||

5. Give the solutions of the equation.

1. Isolate the tangent ratio.
   ||\begin{align}5\tan\big(\!-4(x+1)\big)+2&=6\\5\tan\big(\!-4(x+1)\big)&=4\\\tan\big(\!-4(x+1)\big)&=\dfrac{4}{5}\end{align}||

2. Determine the trigonometric angle.

We determine the angle sought using |\arctan.|||\begin{align}\tan\big(\!-4(x+1)\big)&=\dfrac{4}{5}\\&\Downarrow\\\boldsymbol{\color{#fa7921}{-4(x+1)}}&=\arctan\left(\dfrac{4}{5}\right)\\[3pt]&\approx\boldsymbol{\color{#fa7921}{0.67}}\end{align}||

3. Solve the equation.

We get the following equation that is formed from the angle found in the last step, and solve it.||\begin{align}-4(x+1)&\approx0.67\\x+1&\approx-0.17\\\boldsymbol{\color{#fa7921}{x_1}}&\approx\boldsymbol{\color{#fa7921}{-1.17}}\end{align}||

4. Calculate the period of the tangent function.

The tangent function is periodic, so we need to calculate the period in order to determine all the solutions.||\begin{align}p&=\dfrac{\pi}{\vert b\vert}\\[3pt]&=\dfrac{\pi}{\vert-4\vert}\\[3pt]&=\dfrac{\pi}{4}\end{align}||

5. Give the solutions of the equation.

Since we're looking for solutions within the interval |\left[-\dfrac{3\pi}{2},-\dfrac{\pi}{2}\right],| there are a finite number of solutions. The other solutions are calculated by adding or subtracting the period |\left(\dfrac{\pi}{4}\right)| to the value found in Step 3, without exceeding the interval.

We reject |-1.17,| since this value is larger than |-\dfrac{\pi}{2}.|||\begin{align}x&\approx-1.17-\dfrac{\pi}{4}\\[3pt]&\approx-1.95\end{align}||We keep |-1.95,| since this value is found inside the interval |\left[-\dfrac{3\pi}{2},-\dfrac{\pi}{2}\right].|||\begin{align}x&\approx-1.95-\dfrac{\pi}{4}\\[3pt]&\approx-2.74\end{align}||We keep |-2.74,| since this value is found inside the interval |\left[-\dfrac{3\pi}{2},-\dfrac{\pi}{2}\right].|||\begin{align}x&\approx-2.74-\dfrac{\pi}{4}\\[3pt]&\approx-3.52\end{align}||We keep |-3.52,| since this value is found inside the interval |\left[-\dfrac{3\pi}{2},-\dfrac{\pi}{2}\right].||\begin{align}x&\approx-3.52-\dfrac{\pi}{4}\\[3pt]&\approx-4.31\end{align}||We keep |-4.31,| since this value is found inside the interval |\left[-\dfrac{3\pi}{2},-\dfrac{\pi}{2}\right].|||\begin{align}x&\approx-4.31-\dfrac{\pi}{4}\\[3pt]&\approx-5.1\end{align}||We reject |-5.1,| since this value is less than |-\dfrac{3\pi}{2}.|

A tangent equation of this type can be solved using the same strategies we use to solve a 2nd degree polynomial equation. In other words, you can use factoring or the quadratic formula.

To do so, we need to change the variable by replacing |\tan(x-2)| with |z.| This way, we can temporarily set aside the tangent ratios and concentrate on solving the 2nd degree polynomial.

The result is the following equation:||2\tan^2(x-2)-5\tan(x-2)+2=0\\\Updownarrow\\2z^2-5z+2=0||Using the quadratic formula, we obtain the following solutions for |z|:||\begin{align}z_{1,2}&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\[3pt]&=\dfrac{-(-5)\pm\sqrt{(-5)^2-4(2)(2)}}{2(2)}\\[3pt]&=\dfrac{5\pm\sqrt{9}}{4}\\\\z_1&=\dfrac{1}{2}\quad\text{et}\quad z_2=2\end{align}||The solutions of the equation |2z^2-5z+2=0| are therefore |z_1=\dfrac{1}{2}| and |z_2=2.| Since we changed variables, we can substitute |z| with |\tan(x-2).| We get the following 2 equations:

We can now proceed as we would for a 1st degree tangent equation.

1. Isolate the tangent ratio.

The tangent ratio is already isolated in both equations.

2. Determine the trigonometric angles.

We determine the 2 angles sought using |\arctan| in both equations.

3. Solve the equations.

We get the following 2 equations, which are formed using the angles found in the last step, and solve them.

4. Calculate the period of the tangent function.

The tangent function is periodic, so we need to calculate the period in order to determine all the solutions.||\begin{align}p&=\dfrac{\pi}{\vert b\vert}\\[3pt]&=\dfrac{\pi}{\vert1\vert}\\[3pt]&=\pi\end{align}||

5. Give the solutions to the equation.

1. Change the inequality symbol to an equal symbol.
   ||\begin{align}3\tan\left(\dfrac{x}{4}\right)&>-\sqrt{3}\\&\downarrow\\3\tan\left(\dfrac{x}{4}\right)&=-\sqrt{3}\end{align}||

2. Isolate the tangent ratio.
   ||\begin{align}3\tan\left(\dfrac{x}{4}\right)&=-\sqrt{3}\\[3pt]\tan\left(\dfrac{x}{4}\right)&=-\dfrac{\sqrt{3}}{3}\end{align}||

3. Determine the trigonometric angles.

4. Solve the equations.
   ||\begin{align}\tan\left(\dfrac{x}{4}\right)&=-\dfrac{\sqrt{3}}{3}\\ &\Downarrow\\\dfrac{x}{4}&=-\dfrac{\pi}{6}\\[3pt]\boldsymbol{\color{#fa7921}{x_1}}&=\boldsymbol{\color{#fa7921}{-\dfrac{2\pi}{3}}}\end{align}||

5. Calculate the period of the tangent function.

The tangent function is periodic, so we need to calculate the period in order to determine all the solutions.||\begin{align}p&=\dfrac{\pi}{\vert b\vert}\\[3pt]&=\dfrac{\pi}{\vert\frac{1}{4}\vert}\\[3pt]&=4\pi\end{align}||

6. Calculate the asymptotes of the tangent function.

We calculate the asymptotes on either side of |\boldsymbol{\color{#fa7921}{x_1}}.|

7. Give the solution set of the inequality.

There are 2 possible intervals: either between the lower asymptote |\boldsymbol{(\color{#333fb1}{x_0})}| and |\boldsymbol{\color{#fa7921}{x_1}},| or between |\boldsymbol{\color{#fa7921}{x_1}}| and the upper asymptote |\boldsymbol{(\color{#ec0000}{x_2})}.| To determine which interval is part of the solution set, we can use the graph or test an |x| value in each interval.

Note: Since the inequality symbol is |>,| the boundary associated with |\boldsymbol{\color{#fa7921}{x_1}}| is excluded from the solution set. Moreover, any boundary associated with an asymptote is always excluded.

1. Change the inequality symbol to an equal symbol.
   ||\begin{align}\dfrac{1}{4}\tan\big(2(x-\pi)\big)-6&\le-5\\&\downarrow\\\dfrac{1}{4}\tan\big(2(x-\pi)\big)-6&=-5\end{align}||

2. Isolate the tangent ratio.
   ||\begin{align}\dfrac{1}{4}\tan\big(2(x-\pi)\big)-6&=-5\\[3pt]\dfrac{1}{4}\tan\big(2(x-\pi)\big)&=1\\[3pt]\tan\big(2(x-\pi)\big)&=4\end{align}||

3. Determine the trigonometric angles.

We determine the angle sought using |\arctan.|||\begin{align}\tan\big(2(x-\pi)\big)&=4\\&\Downarrow\\\boldsymbol{\color{#fa7921}{2(x-\pi)}}&=\arctan\left(4\right)\\&\approx\boldsymbol{\color{#fa7921}{1.33}}\end{align}||

4. Solve the equation.
   ||\begin{align}2(x-\pi)&\approx1.33\\x-\pi&\approx0.66\\\boldsymbol{\color{#fa7921}{x_1}}&\approx\boldsymbol{\color{#fa7921}{3.8}}\end{align}||

5. Calculate the period of the tangent function.

The tangent function is periodic, so we need to calculate the period in order to determine all the solutions.||\begin{align}p&=\dfrac{\pi}{\vert b\vert}\\[3pt]&=\dfrac{\pi}{\vert2\vert}\\[3pt]&=\dfrac{\pi}{2}\end{align}||

6. Calculate the asymptotes of the tangent function.

We calculate the asymptotes on either side of |\boldsymbol{\color{#fa7921}{x_1}}.|

7. Give the solution set of the inequality.

There are 2 possible intervals: either between the lower asymptote |\boldsymbol{(\color{#333fb1}{x_0})}| and |\boldsymbol{\color{#fa7921}{x_1}},| or between |\boldsymbol{\color{#fa7921}{x_1}}| and the upper asymptote |\boldsymbol{(\color{#ec0000}{x_2})}.| To determine which interval is part of the solution set, we can use the graph or test an |x| value in each interval.

Note: Since the inequality symbol is |\le,| the boundary associated with |\boldsymbol{\color{#fa7921}{x_1}}| is included in the solution set. Moreover, any boundary associated with an asymptote is always excluded.

1. Change the inequality symbol to an equal symbol.
   ||\begin{align}\tan^2\left(\dfrac{2x}{3}\right)&\ge\dfrac{16}{9}\\&\downarrow\\\tan^2\left(\dfrac{2x}{3}\right)&=\dfrac{16}{9}\end{align}||

2. Isolate the tangent ratio.
   ||\begin{align}\tan^2\left(\dfrac{2x}{3}\right)&=\dfrac{16}{9}\\[3pt]\tan\left(\dfrac{2x}{3}\right)&=\pm\sqrt{\dfrac{16}{9}}\\[3pt]\tan\left(\dfrac{2x}{3}\right)&=\pm\dfrac{4}{3}\end{align}||

3. Determine the trigonometric angles.

We determine the 2 angles sought by using |\arctan| in both equations.

4. Solve the equations.

We get the following equations, formed using the angles found in the last step, and solve them.

5. Calculate the period of the tangent function.

The tangent function is periodic, so we need to calculate the period in order to determine all the solutions.||\begin{align}p&=\dfrac{\pi}{\vert b\vert}\\[3pt]&=\dfrac{\pi}{\left\vert\frac{2}{3}\right\vert}\\[3pt]&=\dfrac{3\pi}{2}\end{align}||

6. Calculate the asymptotes of the tangent function.

We calculate the asymptotes on either side of |\boldsymbol{\color{#fa7921}{x_1}}| and |\boldsymbol{\color{#51b6c2}{x_2}}.|

7. Give the solution set of the inequality.

There are 3 possible intervals: between the lower asymptote |\boldsymbol{(\color{#333fb1}{x_0})}| and |\boldsymbol{\color{#fa7921}{x_1}};| between |\boldsymbol{\color{#fa7921}{x_1}}| and |\boldsymbol{\color{#51b6c2}{x_2}};| and between |\boldsymbol{\color{#51b6c2}{x_2}}| and the upper asymptote |\boldsymbol{(\color{#ec0000}{x_3})}.| To determine which interval is part of the solution set, we can use the graph or test an |x| value in each interval.

Note: Since the inequality symbol is |\ge,| the boundaries associated with |\boldsymbol{\color{#fa7921}{x_1}}| and |\boldsymbol{\color{#51b6c2}{x_2}}| are included in the solution set. Moreover, any boundary associated with an asymptote is always excluded.