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

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

where the quantity $$R$$ and the angle $$\varphi$$ are defined by the following expressions

Tangent Half-Angle Substitution

To solve a non-homogeneous linear equation, we can also use the tangent half-angle formulas:

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

Click or tap a problem to see the solution.

Example 2

Solve the trigonometric equation

$\sin x + \sqrt 3 \cos x = \sqrt 2 .$

Example 2

Solve the trigonometric equation

$\sin x + \cos x = 1.$

Example 3

Solve the equation

$2\sin x + 5\cos x = 8.$

Example 4

Solve the equation

$3\sin 2x + 2\cos 2x = 3.$

Example 5

Solve the equation

$4\sin 3x + \frac{1}{3}\cos 3x = 3.$

Example 6

Find all solutions of the equation in the interval $$\left[ {0,2\pi } \right):$$

$\sqrt 3 \sin \frac{x}{2} - \cos \frac{x}{2} = 1.$

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

$\tan \frac{x}{2} = {t_2} = 1, \Rightarrow \frac{x}{2} = \frac{\pi }{4} + \pi k, \Rightarrow {x_2} = \frac{\pi }{2} + 2\pi k,\,k \in \mathbb{Z}.$

${x_1} = 2\pi n,\;{x_2} = \frac{\pi }{2} + 2\pi k,\;n,k \in \mathbb{Z}.$

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:

$\tan x = {t_1} = \frac{1}{5}, \Rightarrow {x_1} = \arctan \frac{1}{5} + \pi n,\,n \in \mathbb{Z};$
$\tan x = {t_2} = 1, \Rightarrow {x_2} = \frac{\pi }{4} + \pi k,\,k \in \mathbb{Z}.$

${x_1} = \arctan \frac{1}{5} + \pi n,\;{x_2} = \frac{\pi }{4} + \pi k,\;n,k \in \mathbb{Z}.$

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:

$\tan \frac{{3x}}{2} = {t_1} = \frac{2}{5}, \Rightarrow \frac{{3x}}{2} = \arctan \frac{2}{5} + \pi n, \Rightarrow {x_1} = \frac{2}{3}\arctan \frac{2}{5} + \frac{{2\pi n}}{3},\,n \in \mathbb{Z};$
$\tan \frac{{3x}}{2} = {t_2} = 2, \Rightarrow \frac{{3x}}{2} = \arctan 2 + \pi k, \Rightarrow {x_2} = \frac{2}{3}\arctan 2 + \frac{{2\pi k}}{3},\;k \in \mathbb{Z}.$

${x_1} = \frac{2}{3}\arctan \frac{2}{5} + \frac{{2\pi n}}{3},\;{x_2} = \frac{2}{3}\arctan 2 + \frac{{2\pi k}}{3},\;n,k \in \mathbb{Z}.$

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 $$\left[ {0,2\pi } \right):$$ $$x = \frac{{2\pi }}{3}.$$