Question

Solve:tan^2(x)+2sec^2(x)=3

Answer 1

Given:

`\tan ^2\mleft(x\mright)+2\sec ^2\mleft(x\mright)=3`

1. Use the following identity:

`\sec ^2(x)-\tan ^2(x)=1\rightarrow\sec ^2(x)=1+\tan ^2(x)`

This way,

`\begin{gathered} \tan ^2(x)+2\sec ^2(x)=3 \\ \rightarrow\tan ^2(x)+2\lbrack1+\tan ^2(x)\rbrack=3 \\ \rightarrow\tan ^2(x)+2+2\tan ^2(x)=3 \\ \rightarrow2+3\tan ^2(x)=3 \end{gathered}`

2. Clear tan(x) :

`\begin{gathered} 2+3\tan ^2(x)=3 \\ \rightarrow3\tan ^2(x)=1 \\ \rightarrow\tan ^2(x)=(1)/(3)\rightarrow\tan (x)=\pm\frac{1}{\sqrt[]{3}} \\ \rightarrow\tan (x)=\pm\frac{\sqrt[]{3}}{3} \end{gathered}`

Now we know the value of the tangent of the angle we're looking for.

Tan(x) is positive for angles between 0° and 90°, and for angles between 180° and 270°. Knowing this, we need the angles in those intervals that have the tangent we've just calculated. This way, we get that.

`\begin{gathered} x_1=30 \\ x_2=210 \end{gathered}`

Tan(x) is negative for angles between 90° and 180°, and for angles between 270° and 360°. Knowing this, we need the angles in those intervals that have the tangent we've just calculated. This way, we get that.

`\begin{gathered} x_3=150 \\ x_4=330_{} \end{gathered}`

Therefore, the solutions are:

`\begin{gathered} x_1=30 \\ x_2=210 \\ x_3=150 \\ x_4=330_{} \end{gathered}`

Note:

`\begin{gathered} \text{This solutions are for} \\ x\in\lbrack0,360\rbrack \end{gathered}`