Linear Trigonometric Equations
On the previous page, we saw linear homogeneous trigonometric equations of the form
\[a\sin x + b\cos x = 0.\]
Now we consider linear non-homogeneous trigonometric equations in sine and cosine. They are written as
\[a\sin x + b\cos x = c\]
where \(a, b, c \ne 0.\)
Such equations can be solved in several ways. We will deal with two methods here - the \(R\) method and the tangent half-angle substitution.
R Method
The \(R\) method is a technique for converting a linear combination of sine and cosine into a sine (or cosine) function.
Let us derive the \(R\)-formula. For simplicity, let us assume that the coefficients \(a,b\) are positive. The linear form can be represented as follows:
\[a\sin x + b\cos x = \sqrt {{a^2} + {b^2}} \left( {\sin x \cdot \frac{a}{{\sqrt {{a^2} + {b^2}} }} + \cos x \cdot \frac{b}{{\sqrt {{a^2} + {b^2}} }}} \right).\]
We introduce an auxiliary angle \(\varphi\) such that
\[\cos \varphi = \frac{a}{{\sqrt {{a^2} + {b^2}} }},\;\sin \varphi = \frac{b}{{\sqrt {{a^2} + {b^2}} }}.\]
From this it follows that
\[\varphi = \arctan \frac{b}{a}.\]
With the help of the sine addition formula, we get
\[a\sin x + b\cos x = \sqrt {{a^2} + {b^2}} \left( {\sin x\cos \varphi + \cos x\sin \varphi } \right) = \sqrt {{a^2} + {b^2}} \sin \left( {x + \varphi } \right) = R\sin \left( {x + \varphi } \right),\]
where \(R = \sqrt {{a^2} + {b^2}} .\)
Similarly, the difference of sine and cosine can be converted into one trigonometric function by the formula
\[a\sin x - b\cos x = \sqrt {{a^2} + {b^2}} \left( {\sin x\cos \varphi - \cos x\sin \varphi } \right) = \sqrt {{a^2} + {b^2}} \sin \left( {x - \varphi } \right) = R\sin \left( {x - \varphi } \right).\]
The combined \(R-\)formula is therefore
\[a\sin x \pm b\cos x = R\sin\left({x \pm \varphi}\right)\]
where the quantity \(R\) and the angle \(\varphi\) are defined by the following expressions
\[R = \sqrt{a^2 + b^2},\;\;\varphi = \arctan\frac{b}{a}\]
Tangent Half-Angle Substitution
To solve a non-homogeneous linear equation, we can also use the tangent half-angle formulas:
\[\sin x = \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}\]
\[\cos x = \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}\]
Such substitution restricts the domain of the equation:
\[\cos \frac{x}{2} \ne 0, \Rightarrow \frac{x}{2} \ne \frac{\pi }{2} + \pi n, \Rightarrow x \ne \pi + 2\pi n,\,n \in \mathbb{Z}.\]
However, the original linear equation is defined for all real angles \(x \in \mathbb{R}.\) So, you always have to check if \(x = \pi + 2\pi n \) is a solution to the equation.
With the tangent half-angle substitution, a linear equation in sine and cosine can be converted into a rational expression involving only the tangent function of a half-angle.
Solved Problems
Example 1.
Solve the trigonometric equation
\[\sin x + \sqrt 3 \cos x = \sqrt 2 .\]
Solution.
Let's solve this equation using the \(R-\)method. We can easily see that \(a = 1,\) \(b = \sqrt{3}.\) Calculate the number \(R\) and angle \(\varphi:\)
\[R = \sqrt {{a^2} + {b^2}} = \sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}} = 2;\]
\[\varphi = \arctan \frac{b}{a} = \arctan \sqrt 3 = \frac{\pi }{3}.\]
Then the left-hand side \(\left( {LHS} \right)\) of the equation can be written as
\[LHS = \sin x + \sqrt 3 \cos x = 2\sin \left( {x + \frac{\pi }{3}} \right).\]
We got a basic equation which has the following solution:
\[2\sin \left( {x + \frac{\pi }{3}} \right) = \sqrt 2 , \Rightarrow \sin \left( {x + \frac{\pi }{3}} \right) = \frac{{\sqrt 2 }}{2}, \Rightarrow x + \frac{\pi }{3} = {\left( { - 1} \right)^n}\frac{\pi }{4} + \pi n, \Rightarrow x = {\left( { - 1} \right)^n}\frac{\pi }{4} - \frac{\pi }{3} + \pi n,\,n \in \mathbb{Z}.\]
Example 2.
Solve the trigonometric equation
\[\sin x + \cos x = 1.\]
Solution.
First, let us make sure that \(x = \pi + 2\pi n\) is not a solution:
\[LHS = \sin \pi + \cos \pi = 0 + ( - 1) = - 1 \ne RHS = 1.\]
Using the tangent half-angle substitution, we rewrite the equation in the form
\[\frac{{2\tan \frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}} + \frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}} = 1.\]
Let \(\tan \frac{x}{2} = t.\) Simplify the quadratic equation and find its roots:
\[\frac{{2t}}{{1 + {t^2}}} + \frac{{1 - {t^2}}}{{1 + {t^2}}} = 1, \Rightarrow \frac{{2t + 1 - {t^2}}}{{1 + {t^2}}} = 1, \Rightarrow 2t + \cancel{1} - {t^2} = \cancel{1} + {t^2}, \Rightarrow 2{t^2} - 2t = 0, \Rightarrow 2t\left( {t - 1} \right) = 0, \Rightarrow {t_{1,2}} = 0,1.\]
When \(t_1 = 0,\) we have
\[\tan \frac{x}{2} = {t_1} = 0, \Rightarrow \frac{x}{2} = \pi n, \Rightarrow {x_2} = 2\pi n,\,n \in \mathbb{Z}.\]
When \(t_2 = 1,\) the solution is given by
\[\tan \frac{x}{2} = {t_2} = 1, \Rightarrow \frac{x}{2} = \frac{\pi }{4} + \pi k, \Rightarrow {x_2} = \frac{\pi }{2} + 2\pi k,\,k \in \mathbb{Z}.\]
The complete answer is therefore
\[{x_1} = 2\pi n,\;{x_2} = \frac{\pi }{2} + 2\pi k,\;n,k \in \mathbb{Z}.\]
Example 3.
Solve the equation
\[2\sin x + 5\cos x = 8.\]
Solution.
Since sine and cosine cannot be greater than \(1,\) the left-hand side of the equation cannot be greater than \(7.\) Hence,
\[LHS = 2\sin x + 5\cos x \lt 8 = RHS.\]
Therefore, the equation has no solution: \(x \in \varnothing.\)
Example 4.
Solve the equation
\[3\sin 2x + 2\cos 2x = 3.\]
Solution.
Applying the tangent half-angle substitution, we obtain:
\[\frac{{6\tan x}}{{1 + {{\tan }^2}x}} + \frac{{2\left( {1 - {{\tan }^2}x} \right)}}{{1 + {{\tan }^2}x}} = 3, \Rightarrow \frac{{6\tan x + 2 - 2{{\tan }^2}x}}{{1 + {{\tan }^2}x}} = 3.\]
Verify that \(x = \frac{\pi }{2} + \pi n\) is not a solution of the original equation:
\[LHS = 3\sin \left[ {2\left( {\frac{\pi }{2} + \pi n} \right)} \right] + 2\cos \left[ {2\left( {\frac{\pi }{2} + \pi n} \right)} \right] = 3\sin \left( {\pi + 2\pi n} \right) + 2\cos \left( {\pi + 2\pi n} \right) = 0 - 2 = - 2 \ne RHS = 3.\]
Denote \(\tan x = t\) and solve the quadratic equation for \(t.\)
\[\frac{{6t + 2 - 2{t^2}}}{{1 + {t^2}}} = 2, \Rightarrow 6t + 2 - 2{t^2} = 3 + 3{t^2}, \Rightarrow 5{t^2} - 6t + 1 = 0.\]
\[D = {\left( { - 6} \right)^2} - 4 \cdot 5 \cdot 1 = 16 = {4^2}, \Rightarrow {t_{1,2}} = \frac{{6 \pm 4}}{{10}} = \frac{1}{5},1.\]
So there are two possible cases here:
\[\tan x = {t_1} = \frac{1}{5}, \Rightarrow {x_1} = \arctan \frac{1}{5} + \pi n,\,n \in \mathbb{Z};\]
\[\tan x = {t_2} = 1, \Rightarrow {x_2} = \frac{\pi }{4} + \pi k,\,k \in \mathbb{Z}.\]
The answer is given by
\[{x_1} = \arctan \frac{1}{5} + \pi n,\;{x_2} = \frac{\pi }{4} + \pi k,\;n,k \in \mathbb{Z}.\]
Example 5.
Solve the equation
\[4\sin 3x + \frac{1}{3}\cos 3x = 3.\]
Solution.
Let us solve this equation using the tangent half-angle substitution. With this substitution we can lose the root
\[3x = \pi + 2\pi n, \Rightarrow x = \frac{\pi }{3} + \frac{{2\pi n}}{3}.\]
Therefore, we should check that these points are not solutions:
\[LHS = 4\sin \left[ {3\left( {\frac{\pi }{3} + \frac{{2\pi n}}{3}} \right)} \right] + \frac{1}{3}\cos \left[ {3\left( {\frac{\pi }{3} + \frac{{2\pi n}}{3}} \right)} \right] = 3\sin \left( {\pi + 2\pi n} \right) + \frac{1}{3}\cos \left( {\pi + 2\pi n} \right) = 0 - \frac{1}{3} = - \frac{1}{3} \ne RHS = 3.\]
Introduce the new variable \(t = \tan \frac{{3x}}{2},\) so that the sine and cosine functions are written as follows:
\[\sin 3x = \frac{{2\tan \frac{{3x}}{2}}}{{1 + {{\tan }^2}\frac{{3x}}{2}}} = \frac{{2t}}{{1 + {t^2}}},\;\cos 3x = \frac{{1 - {{\tan }^2}\frac{{3x}}{2}}}{{1 + {{\tan }^2}\frac{{3x}}{2}}} = \frac{{1 - {t^2}}}{{1 + {t^2}}}.\]
The original equation then has the form
\[\frac{{8t}}{{1 + {t^2}}} + \frac{{1 - {t^2}}}{{3\left( {1 + {t^2}} \right)}} = 3, \Rightarrow 24t + 1 - {t^2} = 9 + 9{t^2}, \Rightarrow 10{t^2} - 24t + 8 = 0,\]
or
\[5{t^2} - 12t + 4 = 0.\]
The roots of the quadratic equation are \({t_{1,2}} = \frac{2}{5},2.\) So, we have two possible solutions:
\[\tan \frac{{3x}}{2} = {t_1} = \frac{2}{5}, \Rightarrow \frac{{3x}}{2} = \arctan \frac{2}{5} + \pi n, \Rightarrow {x_1} = \frac{2}{3}\arctan \frac{2}{5} + \frac{{2\pi n}}{3},\,n \in \mathbb{Z};\]
\[\tan \frac{{3x}}{2} = {t_2} = 2, \Rightarrow \frac{{3x}}{2} = \arctan 2 + \pi k, \Rightarrow {x_2} = \frac{2}{3}\arctan 2 + \frac{{2\pi k}}{3},\;k \in \mathbb{Z}.\]
The complete answer is
\[{x_1} = \frac{2}{3}\arctan \frac{2}{5} + \frac{{2\pi n}}{3},\;{x_2} = \frac{2}{3}\arctan 2 + \frac{{2\pi k}}{3},\;n,k \in \mathbb{Z}.\]
Example 6.
Find all solutions of the equation in the interval \(\left[ {0,2\pi } \right):\)
\[\sqrt 3 \sin \frac{x}{2} - \cos \frac{x}{2} = 1.\]
Solution.
We will use the \(R-\)formula:
\[a\sin x - b\cos x = R\sin \left( {x - \varphi } \right).\]
Calculate \(R\) and \(\varphi:\)
\[R = \sqrt {{a^2} + {b^2}} = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {1^2}} = 2;\]
\[\varphi = \arctan \frac{b}{a} = \arctan \frac{1}{{\sqrt 3 }} = \frac{\pi }{6}.\]
Then we get the following basic trigonometric equation:
\[2\sin \left( {\frac{x}{2} - \frac{\pi }{6}} \right) = 1, \Rightarrow \sin \left( {\frac{x}{2} - \frac{\pi }{6}} \right) = \frac{1}{2}.\]
There are two sets of solutions to the equation:
\[\frac{x}{2} - \frac{\pi }{6} = {\left( { - 1} \right)^n}\arcsin \frac{1}{2} + \pi n = {\left( { - 1} \right)^n}\frac{\pi }{6} + \pi n, \Rightarrow \frac{x}{2} - \frac{\pi }{6} = \left[ {\begin{array}{*{20}{c}}
{\frac{\pi }{6} + 2\pi n}\\
{\frac{{5\pi }}{6} + 2\pi n}
\end{array}} \right., \Rightarrow \frac{x}{2} = \left[ {\begin{array}{*{20}{c}}
{\frac{\pi }{3} + 2\pi n}\\
{\pi + 2\pi n}
\end{array}} \right., \Rightarrow x = \left[ {\begin{array}{*{20}{c}}
{\frac{{2\pi }}{3} + 4\pi n}\\
{2\pi + 4\pi n}
\end{array}} \right.,\]
where \(n \in \mathbb{Z}.\) There is only one solution in the interval \(\left[ {0,2\pi } \right):\) \(x = \frac{{2\pi }}{3}.\)
