Special Cases for Trigonometric Roots

Introduction To Trigonometric Equations And Roots

An equation involving trigonometric functions such as sine, cosine, tangent, secant, cosecant, cotangent is termed as a trigonometric equation. Solving a trigonometric equation is the process of finding the value of the angles, which satisfies the given equation. There exist two cases while solving a trigonometric equation namely:

  • When the interval is given: The angle satisfying the equation in the given interval is to be found.
  • When no interval is given: The general solution is to be calculated.

Periodic nature indicates that there exist many values, which satisfy the given equation.

Periodic Variation of Trig Functions

The concept of root loss in a trigonometric equation refers to the missing of the roots during the application of algebraic operations. It usually occurs while the similar terms are cancelled out on both sides. It is explained using the following example.

Let sin θ [A] = sin θ [B]. Here, A and B can be termed as trigonometric functions. Upon cancelling sin θ, the solutions are lost. sin θ = 0 is obtained.

i.e. θ = n π, where n = 0, ±1, ±2, ………

Thus, the cancellation can’t be done directly.

Extraneous roots are the ones which are obtained while the terms in trigonometric equations are squared. The check for extraneous roots must be done in order to obtain the solution accurately.

The steps to solve trigonometric equations are as follows:

  • The given equation is taken in the form of a function of an angle.
  • The equation, written as a trigonometric function of an angle is constant.
  • The possible angles for the angle are listed out.
  • The variable is solved.
  • The restriction is applied to the solution.

Special Cases For Trigonometric Roots Examples

Example 1: The number of pairs (x, y) satisfying the equations sin x + sin y = sin (x + y) and |x|+ |y| = 1 is ____________.

Solution:

The first equation can be written as 2 sin [1 / 2] (x + y) cos [1 / 2] (x − y)

= 2 sin [1 / 2] (x + y) cos [1 / 2] (x + y)

∴ Either sin [1 / 2] (x + y) = 0 or sin [1 / 2] x = 0 or sin [1 / 2] y = 0

Thus, x + y = −1, x − y = −1.

When x + y = 0, we have to reject x + y = 1 and check with the options or x + y = −1 and solve it with x − y = 1 or x − y = −1 which gives (1 / 2, −1 / 2) or (−1 / 2, 1 / 2) as the possible solution.

Again, solving with x = 0, we get (0, ±1) and solving with y = 0, we get (± 1, 0) as the other solution. Thus, we have six pairs of solutions for −4, −3, −2, −1, 0, 1, 2, 3 and y.

Example 2: The general solution of sin x − 3 sin 2x + sin 3x = cos x − 3 cos 2x + cos 3x is __________.

Solution:

sin x − 3 sin 2x + sin 3x = cos x − 3 cos 2x + cos 3x

⇒ 2 sin2x cosx − 3 sin2x − 2 cos2x cosx + 3 cos2x = 0

⇒ sin2x (2cosx − 3) − cos2x (2cosx − 3) = 0

⇒ (sin2x − cos2x) (2cosx − 3) = 0

⇒ sin2x = cos2x

⇒ 2x = 2nπ ± (π / 2 − 2x)

i.e., x = [nπ / 2] + [π / 8]

Example 3: Find the solution to the following equation such that 0 ≤ x < 2π, 2sin2 (x) − sin (x) + 3 = 4

Solution:

2 sin2(x) − sinx − 1 = 0

Next, we factor.

(2 sin (x) + 1) (sin (x) − 1) = 0

Set each factor equal to zero and solve.

2 sin (x) + 1 = 0 or sin (x) − 1 = 0

sin (x) = −1 / 2 sin (x) = 1

Determine the angles that satisfy each solution within one revolution.

The angles 7π / 6 and 11π / 6 satisfy the first, and π / 2 satisfies the second.

Example 4: Solve the following equation for 0 ≤ x < 2π, sin 2x = cos 2x

Solution:

Let u = 2x.

The equation now becomes:

sin u = cos u

The sine and cosine functions are equal at the following angles.

u = π / 4; 5π / 4; 9π / 4; 13π / 4

These angles are divided by 2 so that it is within the range.

2x = π / 4; 2x = 5π / 4; 2x = 9π / 4; 2x = 13π / 4

Dividing each answer by 2 gives

x = π / 8; 5π / 8; 9π / 8; 13π / 8

Example 5: Solve the equation for 0 ≤ x < 2π, 3 tan2 x = 1

Solution:

3 tan2 x = 1;

Divide both sides by 3

tan2 x = 1 / 3

Take the square root on both sides.

tan x = ± 1 / √3

This can be written as two separate equations

tan x = 1 / √3 and tan x = −1 / √3

Take the inverse tangent

x = π / 6; 7π / 6 and x = 5π/ 6; 11π / 6

Example 6: If cos θ + cos 7θ + cos 3θ +cos 5θ = 0, then what is the value of θ?

Solution:

Combining θ and 7θ, 3θ and 5θ, we get

2 cos 4θ (cos 3θ + cos θ) = 0

4 cos 4θ cos 2θ cosθ = 0

4 [(1 / 23) sinθ] (sin23 θ) = 0; sin 8θ = 0.

Hence, θ = nπ / 8

Example 7: If n is any integer, then the general solution of the equation cosxsinx=12\cos x-\sin x=\frac{1}{\sqrt{2}} is

A)x=2nππ12 or x=2nπ+7π12B)x=nπ±π12C)x=2nπ+π12 or x=2nπ7π12D)x=nπ+π12 or x=nπ7π12A) x=2n\pi -\frac{\pi }{12} \text \ or \ x=2n\pi +\frac{7\pi }{12}\\ B) x=n\pi \pm \frac{\pi }{12}\\ C) x=2n\pi +\frac{\pi }{12} \text \ or \ x=2n\pi -\frac{7\pi }{12}\\ D) x=n\pi +\frac{\pi }{12} \text \ or \ x=n\pi -\frac{7\pi }{12}\\

Solution: 

Given equation is, cosxsinx=12\cos x-\sin x=\frac{1}{\sqrt{2}}

Dividing equation by 2\sqrt{2}

12cosx12sinx=12cos(π4+x)=cosπ3.\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x=\frac{1}{2}\\ \cos \left( \frac{\pi }{4}+x \right)=\cos \frac{\pi }{3}.

Hence, 

π4+x=2nπ±π3x=2nπ+π3π4=2nπ+π12x=2nππ3π4=2nπ7π12.\frac{\pi }{4}+x=2n\pi \pm \frac{\pi }{3}\\ x=2n\pi +\frac{\pi }{3}-\frac{\pi }{4}=2n\pi +\frac{\pi }{12}\\ x=2n\pi -\frac{\pi }{3}-\frac{\pi }{4}=2n\pi -\frac{7\pi }{12}.

Example 8: The value of θ\theta lying between 0 and π/2\pi /2 and satisfying the equation 1+sin2θcos2θ4sin4θsin2θ1+cos2θ4sin4θsin2θcos2θ1+4sin4θ=0\left| \,\begin{matrix} 1+{{\sin }^{2}}\theta & {{\cos }^{2}}\theta & 4\sin 4\theta \\ {{\sin }^{2}}\theta & 1+{{\cos }^{2}}\theta & 4\sin 4\theta \\ {{\sin }^{2}}\theta & {{\cos }^{2}}\theta & 1+4\sin 4\theta \\ \end{matrix}\, \right|=0

Solution:

The given determinant ( Applying R1R1R3 and R2R2R3)(\text \ Applying \ {{R}_{1}}\to {{R}_{1}}-{{R}_{3}} \text \ and \ {{R}_{2}}\to {{R}_{2}}-{{R}_{3}}) reduces to 

101011sin2θcos2θ1+4sin4θ=01+4sin4θ+cos2θ+sin2θ=0\left| \,\begin{matrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ {{\sin }^{2}}\theta & {{\cos }^{2}}\theta & 1+4\sin 4\theta \\ \end{matrix}\, \right|\,=0\\ \Rightarrow 1+4\sin 4\theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =0

(By expanding along R1){{R}_{1}})

4sin4θ=2sin4θ=124θ=7π611π6,(0<4θ<2π)4\sin 4\theta =-2 \\ \sin 4\theta =\frac{-1}{2}\\ 4\theta =\frac{7\pi }{6}\\ \frac{11\pi }{6}, (0<4\theta <2\pi )

Since,

0<θ<π20<4θ<2πθ=7π24,11π240<\theta <\frac{\pi }{2}\\ 0<4\theta <2\pi \\ \theta =\frac{7\pi }{24},\,\,\frac{11\pi }{24}

Example 9: Common roots of the equations 2sin2x+sin22x=22{{\sin }^{2}}x+{{\sin }^{2}}2x=2 and sin2x+cos2x=tanx,\sin 2x+\cos 2x=\tan x, are _____.

Solution: 

2sin2x+sin22x=2(i)sin2x+cos2x=tanx(ii)2{{\sin }^{2}}x+{{\sin }^{2}}2x=2 \rightarrow (i) \\ \sin 2x+\cos 2x=\tan x \rightarrow (ii)

Solving (i), 

sin22x=2cos2x2cos2xcos2x=0x=(2n+1)π2 or x=(2n+1)π4{{\sin }^{2}}2x=2{{\cos }^{2}}x\\ 2{{\cos }^{2}}x\cos 2x=0\\ x=(2n+1)\frac{\pi }{2}\text{ or }x=(2n+1)\frac{\pi }{4}

Therefore, common roots are (2n±1)π4(2n\pm 1)\frac{\pi }{4}

Solving (ii),

2tanx+1tan2x1+tan2x=tanxtan3x+tan2xtanx1=0(tan2x1)(tanx+1)=0x=mπ±π4\frac{2\tan x+1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\tan x\\ \Rightarrow {{\tan }^{3}}x+{{\tan }^{2}}x-\tan x-1=0\\ \Rightarrow ({{\tan }^{2}}x-1)\,(\tan x+1)=0\\ \Rightarrow x=m\pi \pm \frac{\pi }{4}

Example 10: The number of values of x in the interval [0,5π][0, 5 \pi ] satisfying the equation 3sin2x7sinx+2=03{{\sin }^{2}}x-7\sin x+2=0 is 

Solution:

3sin2x7sinx+2=03sin2x6sinxsinx+2=03sin(sinx2)(sinx2)=0(3sinx1)(sinx2)=0sinx=13 or 2sinx=13,3{{\sin }^{2}}x-7\sin x+2=0\\ \Rightarrow 3{{\sin }^{2}}x-6\sin x-\sin x+2=0\\ \Rightarrow 3\sin (\sin x-2)-(\sin x-2)=0\\ \Rightarrow (3\sin x-1)\,(\sin x-2)=0\\ \Rightarrow \sin x=\frac{1}{3}\text{ or 2}\\ \Rightarrow \sin x=\frac{1}{3},

(Because sinx2\sin x \ne 2)

Let sin113=α,0<α<π2{{\sin }^{-1}}\frac{1}{3}=\alpha , 0<\alpha <\frac{\pi }{2} are the solutions in [0, 5π][0,\text{ }5\pi ].

Then α,πα,2π+α,3πα,4π+α,5πα\alpha ,\pi -\alpha ,\,2\pi +\alpha ,\,3\pi -\alpha ,\,4\pi +\alpha , 5\pi -\alpha are the solutions in [0,5π][0,\,5\pi ]

Required number of solutions = 6.

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