Find the general solution to $\csc \theta + \sec \theta = 1$

Question

Find the general solution to $$\csc\theta + \sec\theta =1$$

This is how I solved. We have,

\begin{align} \csc\theta + \sec\theta &=1\\ \frac1{\sin\theta} + \frac1{\cos\theta}& =1\\ \frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta} &=1\\ (\sin\theta + \cos\theta)^2 &= (\sin\theta\cos\theta)^2 \\ 1 + 2\sin\theta\cos\theta &= \frac{4\sin^2\theta\cos^2\theta}4\\ 1 + \sin2\theta &= \frac{(2\sin\theta\cos\theta)^2 }4\\ 4 + 4\sin2\theta &= \sin^2 2\theta\\ \sin^2 2\theta - 4\sin2\theta - 4 &= 0\\ \sin2\theta &= 2 - 2\sqrt2\end{align}

Now here I am stuck. Can someone please help me proceed further?

Answer 1

Your squaring of the equation $$\cos x+\sin x=\cos x\>\sin x\tag{1}$$ has introduced spurious solutions. In fact the value ${1\over2}\arcsin\bigl(2-2\sqrt{2}\bigr)\approx-0.488147$ does not solve the given problem.

Drawing the graphs of $x\mapsto \cos x+\sin x$ and $x\mapsto\cos x\>\sin x$ shows a symmetry with respect to $x={\pi\over4}$. We therefore put $x:={\pi\over4}+t$ and then have $$\cos x+\sin x=\sqrt{2}\>\cos t,\qquad\cos x\>\sin x={1\over2}\cos(2t)\ .$$ Plugging this into $(1)$ we obtain $$\sqrt{2}\cos t={1\over2}(2\cos^2 t-1)\ ,$$ so that $\cos t={\sqrt{2}\over2}-1$, or $$ t=\pm \alpha,\quad{\rm with}\quad \alpha:=\arccos{\sqrt{2}-2\over2}=1.86805\ .$$ This leads to the $x$-values $$x_1={\pi\over4}-\alpha=-1.08265,\qquad x_2={\pi\over4}+\alpha=2.65345\ .$$ Looking at the graphs we see that these solutions repeat with periodicity $2\pi$.

Answer 2

If you use the tangent half-angle substitution $\theta=2 \tan ^{-1}(t)$, the expression write $$\frac{t^4-4 t^3-1}{t(1+t)(1-t)}=0$$ Assuming that the denominator cannot cancel, the real solutions of $t^4-4 t^3-1=0$ are given by $$t_\pm=1+\frac{1}{\sqrt{2}}\pm\sqrt{\frac{1}{2} \left(5+4 \sqrt{2}\right)}$$ which makes the solution to be $$\theta_\pm=2 \tan^{-1}\left(1+\frac{1}{\sqrt{2}}\pm\sqrt{\frac{1}{2} \left(5+4 \sqrt{2}\right)} \right)+n \pi$$ and I do not think that we could do more.

We can do some amazing things considering that it is the same as finding the zero's of $$f(x)=\sin (x) \cos (x) (\csc (x)+\sec (x)-1)=\sin (x)+\cos (x)-\sin (x) \cos (x)$$ The first positive root is close to $\frac {5\pi}6$. Then, using Taylor series $$f(x)=\left(\frac{1}{2}-\frac{\sqrt{3}}{4}\right)-\left(1+\frac{\sqrt{3}}{2}\right) \left(x-\frac{5 \pi }{6}\right)-\left(\frac{1}{4}+\frac{\sqrt{3}}{4}\right) \left(x-\frac{5 \pi }{6}\right)^2+\frac{1}{12} \left(5+\sqrt{3}\right) \left(x-\frac{5 \pi }{6}\right)^3+O\left(\left(x-\frac{5 \pi }{6}\right)^4\right)$$

Now, using series reversion $$x_+=\frac{5 \pi }{6}+\left(2 \sqrt{3}-4\right) \left(y-\frac{1}{4} \left(2-\sqrt{3}\right)\right)+\left(38-22 \sqrt{3}\right) \left(y-\frac{1}{4} \left(2-\sqrt{3}\right)\right)^2+\left(692 \sqrt{3}-\frac{3596}{3}\right) \left(y-\frac{1}{4} \left(2-\sqrt{3}\right)\right)^3+O\left(\left(y-\frac{1}{4} \left(2-\sqrt{3}\right)\right)^4\right)$$ where $y$ stands for $f(x)$. Making $y=0$ leads to $$x_+=\frac{5 \pi }{6}+\frac{48487-27993 \sqrt{3}}{48}\approx 2.6534469$$ while the exact solution, obtained using Newton method, is $2.6534459$

The first neagtive root is close to $-\frac \pi 3$. Doing the same $$f(x)=\left(\frac{1}{2}-\frac{\sqrt{3}}{4}\right)+\left(1+\frac{\sqrt{3}}{2}\right) \left(x+\frac{\pi }{3}\right)-\left(\frac{1}{4}+\frac{\sqrt{3}}{4}\right) \left(x+\frac{\pi }{3}\right)^2-\left(\frac{5}{12}+\frac{1}{4 \sqrt{3}}\right) \left(x+\frac{\pi }{3}\right)^3+O\left(\left(x+\frac{\pi }{3}\right)^4\right)$$

$$x_-=-\frac{\pi }{3}+\left(4-2 \sqrt{3}\right) \left(y-\frac{1}{4} \left(2-\sqrt{3}\right)\right)+\left(22 \sqrt{3}-38\right) \left(y-\frac{1}{4} \left(2-\sqrt{3}\right)\right)^2+\left(\frac{3596}{3}-692 \sqrt{3}\right) \left(y-\frac{1}{4} \left(2-\sqrt{3}\right)\right)^3+O\left(\left(y-\frac{1}{4} \left(2-\sqrt{3}\right)\right)^4\right)$$ Making $y=0$ leads to $$x_-=-\frac \pi 3-\frac{48487-27993\sqrt{3}}{48}\approx -1.0826505 $$while the exact solution, obtained using Newton method, is $-1.0826495$.

Answer 3

Hint

Again avoid squaring as it immediately introduces When do we get extraneous roots?

$\sin x\cos x=\sin x+\cos x=y,$

$y=\sqrt2\cos(x-45^\circ)\implies-\sqrt2\le y\le?$(say)

$$y^2=1+2\sin x\cos x$$

Let $y^2-1=2y\iff y^2-2y-1=0$

$y=1\pm\sqrt2$

$\implies y=1-\sqrt2$

$\implies\cos(x-45^\circ)=\dfrac1{\sqrt2}-1$