Equations and identities

​​In a nutshell

Quadratic equations in $$\sin(\theta)$$​, $$\cos(\theta)$$ or $$\tan(\theta)$$ can be solved by first solving the quadratic in an arbitrary variable, then solving the simple trigonometric equations corresponding to the roots. More generally, you can use the trigonometric identities $$\sin^2(\theta) + \cos^2(\theta) \equiv 1$$​ and $$\dfrac{\sin(\theta)}{\cos(\theta)} \equiv \tan(\theta)$$​ to help you solve a wide range of equations involving multiple trigonometric functions.

Quadratic equations in a single trigonometric function

Suppose you are given an equation in one of the following forms:

$$\begin{aligned}a\sin^2(\theta) + b\sin(\theta) + &c = 0\\a\cos^2(\theta) + b\cos(\theta) + &c = 0\\a\tan^2(\theta) + b\tan(\theta) + &c = 0\end{aligned}$$​

​​

Where $$a$$​, $$b$$​ and $$c$$ are constants. The following procedure allows you to find all solutions to any of the above equations in any interval $$D \leq \theta \leq D'$$:

PROCEDURE

Example 1

Find all solutions to the equation $$\sin^2(\theta) = \dfrac14$$ in the interval $$0\degree \le \theta \le 360\degree$$.

Consider the equation $$x^2 = \dfrac14$$. Rearrange and factorise to get:

$$\begin{aligned}x^2 &= \dfrac14\\x^2 - \dfrac14 &= 0\\\left(x-\dfrac12\right)\left(x+\dfrac12\right) &= 0\end{aligned}$$​​

This means that the equation $$\sin^2(\theta) = \dfrac14$$ can be solved by solving the two equations:

$$\sin(x)=\dfrac{1}{2}$$​​

$$\sin(x)=-\dfrac{1}{2}$$​​

The principal value for the first root is $$\sin^{-1}\left(\dfrac12\right)\degree = 30\degree$$. Using acute angle identities or otherwise, determine that the only solutions to $$\sin(\theta) = \dfrac12$$ in the interval $$0\degree \le \theta \le 360\degree$$ are $$\theta = 30\degree,\space 150\degree$$.

The principal value for the second root is $$\sin^{-1}\left(-\dfrac12\right)\degree = -30\degree$$. Using acute angle identities or otherwise, determine that the only solutions to $$\sin(\theta) = -\dfrac12$$ in the interval $$0\degree \le \theta \le 360\degree$$ are $$\theta = 210\degree,\space \theta = 330\degree$$.

Therefore, the solutions to the equation $$\sin^2(\theta) = \dfrac14$$ in the interval $$0\degree \le \theta \le 360\degree$$ are $$\underline{\theta = 30\degree}$$, $$\underline{\theta = 150\degree}$$, $$\underline{\theta = 210\degree}$$ and $$\underline{\theta = 330\degree}$$.

​​Equations in multiple trigonometric functions

You will often be given equations with multiple different trigonometric functions. Using the identities $$\sin^2(\theta) + \cos^2(\theta) \equiv 1$$​ and $$\dfrac{\sin(\theta)}{\cos(\theta)} \equiv \tan(\theta)$$​,  you will need to rearrange and factorise these equations into simple trigonometric equations you can solve.

Most of the time it will be possible to use these identities to transform the equation into a quadratic equation in a single trigonometric function, which can be solved by applying the above procedure.

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Example 2

Find all solutions to the equation $$\cos(\theta)(\cos(\theta) + \tan(\theta)) = 0$$ in the interval $$0\degree \le \theta \le 360\degree$$. Give your answers to $$3\space s.f.$$​

Use the identities $$\tan(\theta) \equiv \dfrac{\sin(\theta)}{\cos(\theta)}$$ and $$\sin^2(\theta) + \cos^2(\theta) \equiv 1$$ to rearrange the equation so that it is a quadratic in $$\sin(\theta)$$:

$$\begin{aligned}\cos(\theta)\left(\cos(\theta) + \tan(\theta)\right)&\equiv \cos(\theta)\left(\cos(\theta)+\dfrac{\sin(\theta)}{\cos(\theta)} \right)\\&\equiv \cos^2(\theta) + \sin(\theta)\\&\equiv 1 + \sin(\theta) - \sin^2(\theta)\end{aligned}$$​​

Therefore, it suffices to find all solutions to the equation $$\sin^2(\theta) - \sin(\theta) - 1 = 0$$ in the interval $$0\degree \le \theta \le 360\degree$$.

The equation $$x^2 - x - 1=0$$ yields the solutions $$x=\dfrac{1\pm\sqrt{5}}{2}$$.

The equation ​$$\sin(\theta) = \dfrac{1 +\sqrt5}{2}$$ has no solutions, as  $$\dfrac{1 + \sqrt5}{2} > 1$$.

The principal value for the second root is $$\sin^{-1}\left(\left[\dfrac{1-\sqrt5}{2}\right]\right)\degree = -38.17\dots\degree$$. Using acute angle identities or otherwise, determine that the only solutions to the equation $$\sin(\theta) = \dfrac{1-\sqrt5}{2}$$ in the interval $$0\degree \le \theta \le 360\degree$$ are $$\theta = 218.1\dots\degree,\space 321.8\dots\degree$$.

Therefore, to $$3\space s.f.$$, the solutions to $$\cos(\theta)(\cos(\theta) + \tan(\theta)) = 0$$ in the interval $$0 \degree\le \theta \le 360\degree$$ are $$\underline{\theta = 218\degree}$$ and $$\underline{\theta = 322\degree}$$.