What is the general solution of #2tan^2 3x-sec3x=1#?

Question

Brooklyn Anderson · Accepted Answer

#x=120^@xxn+-16.06^@# or #x=120^@xxn+60^@#, where #n# is an integer.

#2tan^2 3x-sec3x=1# can be written as #2(sec^2 3x-1)-sec3x=1# or #2sec^2 3x-sec3x-3=0# or #(2sec3x-3)(sec3x+1)=0# i.e. #sec3x=3/2=sec48.19^@# i.e. #3x=360^@xxn+-48.19^@# i.e. #x=120^@xxn+-16.06^@# or #sec3x=-1=sec180^@# i.e. #3x=360^@xxn+180^@# i.e. #x=120^@xxn+60^@#, where #n# is an integer.

Emery Adkins · Answer

undefined To find the general solution of $2\tan^2(3x) - \sec(3x) = 1$, first, rewrite $\tan^2(3x)$ in terms of $\sec(3x)$, then solve the resulting quadratic equation in $\sec(3x)$.

$$
\begin{align*}
2\tan^2(3x) - \sec(3x) &= 1 \
2\left(\frac{\sec^2(3x) - 1}{\sec^2(3x)}\right) - \sec(3x) &= 1 \
2\left(\frac{1}{\sec^2(3x)} - \frac{1}{\sec(3x)}\right) - \sec(3x) &= 1 \
2\left(\frac{1 - \cos^2(3x)}{\cos^2(3x)}\right) - \sec(3x) &= 1 \
2\left(\frac{\sin^2(3x)}{\cos^2(3x)}\right) - \sec(3x) &= 1 \
2\tan^2(3x) - \sec(3x) &= 1 \
\Rightarrow 2\sin^2(3x) - \cos^2(3x) &= \cos(3x) \
2(1 - \cos^2(3x)) - \cos^2(3x) &= \cos(3x) \
2 - 3\cos^2(3x) &= \cos(3x) \
3\cos^2(3x) + \cos(3x) - 2 &= 0
\end{align*}
$$

Now, solving this quadratic equation in $\cos(3x)$, we find the roots and then substitute them back to find $3x$. Finally, we divide each solution of $3x$ by 3 to find the solutions for $x$.