Simplifying, I have

$$\frac{1}{\sin x\cdot \cos x} = 3$$

I have tried many manipulations but did not get the answer. Point me the right direction to the solution. (This problem is in the beginning section of my text. Only the basic functions, pythagorean identities and the unit circle were introduced up to this point.)

Added: Double angle, half angle, addition and subtraction formulas were not mentioned yet.


Method $1$: Recall that $\sin(2x) = 2\sin(x)\cos(x)$. From what you have we have $$\sin(x) \cos(x) = \dfrac13 \implies \sin(2x) = \dfrac23$$ Since $x$ lies in the first quadrant, $2x$ lies in the first or second quadrant. Hence, we have $$2x = \arcsin\left(\dfrac23\right) \text{ or }\pi-\arcsin\left(\dfrac23\right)$$ This gives us that $$x = \dfrac12 \arcsin\left(\dfrac23\right) \text{ or }\dfrac{\pi}2 - \dfrac12 \arcsin\left(\dfrac23\right)$$

Method $2$: We have $\sin(x) \cos(x) = \dfrac13$. This gives us that $$\sin^2(x) \cos^2(x) = \dfrac19 \implies \sin^2(x)(1-\sin^2(x)) = \dfrac19$$ Setting $\sin^2(x) = t$, we obtain a quadratic in $t$, i.e., $$t(1-t) = \dfrac19 \implies t^2 - t + \dfrac19 =0 \implies \left(t-\dfrac12\right)^2 = \dfrac14 - \dfrac19 = \dfrac5{36} \implies t = \dfrac12 \pm \dfrac{\sqrt5}6$$ Hence, we have $\sin^2(x) = \dfrac{3 \pm \sqrt5}6$. Since we are looking for $x$ in the first quadrant, $\sin(x)$ has to be positive. Hence, we obtain $$x = \arcsin\left(\sqrt{\dfrac{3 \pm \sqrt5}6}\right)$$

  • $\begingroup$ Your method 2 is what I used. +1. $\endgroup$ – MPW Apr 15 '15 at 2:54
  • $\begingroup$ Only sin x was asked,via method 2. Thanks for the detailed solution. :D $\endgroup$ – Aditya Chintalapati Apr 15 '15 at 3:02

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