Precalculus

Trigonometry

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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

Linear nonhomogeneous trig equation

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

Trigonometric R-Formula

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

Expressions for radius R and angle Phi

Tangent Half-Angle Substitution

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

Tangent Half-Angle Formulas for Sine and Cosine

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

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

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