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Solve for $\theta$ in $2\sec^2θ-4 =0$

I have gotten toward $\sec^2θ=2$ Then $\dfrac{1}{\cos^2θ} = 2$

What is the next step to this problem?

Sujaan Kunalan
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Jon
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4 Answers4

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Next Steps: $\cos^2\theta=\dfrac{1}{2}$, then $\cos\theta=\pm\dfrac{1}{\sqrt{2}}$.

Going to cosine was a good idea. Alternately, you could use the identitity $1+\tan^2\theta=\sec^2\theta$ to get $2(\tan^2\theta+1)=4$, and then $\tan^2\theta=1$, then $\tan\theta=\pm 1$.

André Nicolas
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  • Then you get $cosθ= √2/2 = π/4 + 2nπ$ – Jon Aug 02 '13 at 17:14
  • There are a lot more. In the interval $0$ to $2\pi$ there are $4$ values, and then you add $2n\pi$. More compactly, all values are given by $\frac{\pi}{4}+\frac{n\pi}{2}$ or equivalently odd multiples of $\frac{\pi}{4}$. – André Nicolas Aug 02 '13 at 17:45
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$\dfrac{1}{\cos^2\theta}=2 \implies\cos^2 \theta=\dfrac{1}{2}\implies \cos \theta=\pm\dfrac{1}{\sqrt{2}}\implies\theta=\arccos(\pm\dfrac{1}{\sqrt{2}})$

Sujaan Kunalan
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Method $1:$

When we only have $\cos^2\theta$ in the equation, its better to utilize $\cos2\theta=2\cos^2\theta-1$

As we have $2\cos^2\theta-1=0, \cos2\theta=0$

$\displaystyle\implies 2\theta=(2n+1)\frac\pi2$ where $n$ is any integer

or as we know if $\cos x=\cos \alpha,x=2n\pi\pm \alpha$

$\displaystyle\implies 2\theta=2p\pi\pm\frac\pi2$ where $p$ is any integer

Method $2:$

Alternatively, we know if $\cos^2x=\cos^2A\iff \sin^2x=\sin^2A\iff \tan^2x=\tan^2A$

$\displaystyle x=m\pi\pm A$ where $m$ is any integer

Now, we have $\cos^2\theta=\frac12=\left(\frac1{\sqrt2}\right)^2=\cos^2\frac\pi4$

$\implies x=r\pi\pm \frac\pi4$ where $r$ is any integer

Can you show that all the solutions are equivalent?

lab bhattacharjee
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  • how do you decided when to use $πn$ or $2πn$? – Jon Aug 02 '13 at 17:55
  • @LittleJon, we prove $\cos(n\pi+\frac\pi2)=\cos n\pi \cos\frac\pi2 - \sin n\pi \sin\frac\pi2=0$ and $\cos x=1,x=2n\pi$ . For other values if $\cos x=\cos \alpha,x=2n\pi\pm \alpha$ – lab bhattacharjee Aug 02 '13 at 17:57
  • @LittleJon, you must have observed that the last formula can be applied for the values of $\cos\alpha$. But for $\cos x=\pm1,$ two values converge. – lab bhattacharjee Aug 02 '13 at 18:10
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This is a difference of squares:

$0 = 2 \sec^2 \theta - 4 = 2( \sec \theta + \sqrt2)( \sec \theta - \sqrt2)$

So one of the two factors of the product on the right is $0$. In the first case, $\sec \theta = - \sqrt2$, so $\cos \theta = -\frac{1}{\sqrt2}$, and in the second $\cos \theta = \frac{1}{\sqrt2}$. Each of these possibilities has infinitely many solutions for $\theta$ real, but if you restrict to $\theta$ in the interval $[0,\pi]$ then the unique solutions are given by $\theta = \cos^{-1}(\pm \frac{1}{\sqrt{2}})$.

user88377
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