TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 7 Trigonometric Equations to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 1.
Solve 2cos2θ – \(\sqrt{3}\) sin θ+1 = 0
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 1

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 2.
Solve 1 +sin2θ = 3 sinθ cosθ
Solution:
Dividing by cos2 θ we get
⇒ sec2 θ + tan2 θ = 3 tanθ
⇒ 1 + 2 tan2 θ = 3 tanθ
⇒ 2tan2 θ – 3tan θ +1 = θ
⇒ 2 tan2 θ –  2tan θ –  tanθ + 1 = θ
⇒ (tanθ –  1)(2tan θ – 1) = θ
⇒ tanθ = 1 or tanθ = \(\frac{1}{2}\)
If tan θ = 1 then principal solution is α =\( \frac{\pi}{4}\)
General solution is θ = nπ + \(\frac{\pi}{4}\)  n ∈ Z
Let a be the principal solution of tan θ = \(\frac{1}{2}\)
Then the general solution is
θ = nπ+α,n∈ Z

Question 3.
Solve \(\sqrt{2}(\sin x+\cos x)=\sqrt{3}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 3

Question 4.
Solve tanθ+3 cotθ = 5 secθ
Solution:
The given equation is
tanθ + 3 cotθ = 5 secθ
⇒ \(\frac{\sin \theta}{\cos \theta}+\frac{3 \cos \theta}{\sin \theta}=\frac{5}{\cos \theta}\)
⇒  sin2 θ + 3cos2θ= 5 sinθ
⇒  sin2 θ + 3(1 – sin2 θ) = 5 sinθ
⇒ sin2 θ + 3 – 3 sin2 θ = 5 sinθ
⇒ -2 sin θ – 5 sinθ + 3 = 0
⇒ 2 sin θ+ 5 sinθ – 3 = θ
⇒ 2 sin θ + 6 sinθ – sinθ – 3 = θ
⇒ 2 sin θ(sin θ + 3) –  1(sinθ + 3) = θ
⇒ (2sinθ – 1)(sinθ + 3)= θ
⇒ sin θ = \(\frac{1}{2}\) or sinθ = – 3.
sinθ =-3 is not admissible but for sinθ = \(\frac{1}{2}\) general solution is
θ=nπ+(-n)n ,n∈Z
Since the principal solution is α = \(\frac{1}{2}\)

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 5.
If acos2θ + bsin2θ =c has θ12 as its solutions then show that tan θ1 + tan θ1 = \(\frac{2 b}{c+a}\)
Solution:
Given equation is acos2θ + bsin2θ = c
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 4
This is a quadratic equation in an θ.
Given θ12  are the solutions of θ.
tan θ1, tanθ2 are the roots of equation (1)
∴ Sum of the roots tan θ1 + tan θ2 = \(\frac{2 b}{a+c}\) and product of the roots
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 5

Question 6.
Find all values of x in (-π, π) satisfying the equation g1+cosx+cos2 x +……………….. = 43
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 6

Question 7.
Solve sin x = \(\frac{1}{\sqrt{2}}\)
Solution:
Principal solution is α = \(\frac{\pi}{4}\)
General solution is
\(\mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}\)

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 8.
Solve sin 2θ \(=\frac{\sqrt{5}-1}{4}\)
Solution:
The principal solution is α = 18°
∴ General solution is
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 7
Question 9.
Solve tan2 θ = 3.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 8
Question 10.
Solve 3cosec x = 4sinx.
Solution:
Given 3 cosec x = 4 sin x
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 9

Question 11.
If x is acute and sin(x+10°) = cos(3x – 68°) find x in degrees.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 10
When k = 0 we get x = 37°
If we take k = 1, 2 the value of x is not acute.
Hence the value of x is 37°.

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 12.
Solve cos 3θ = sin 2θ.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 11

Question 13.
Solve 7 sin2θ+3cos2 θ= 4.
Solution:
Given 7 sin2θ+3cos2 θ= 4.
⇒ 7 sin 2θ + 3(1 -sin2 θ) = 4
⇒  4 sin 2θ = 1
⇒ sinθ = ± \(\frac{1}{2}\)
Principal solutions are \(\alpha=\pm \frac{\pi}{6}\) and general solution is given by general solution is given by
θ = nπ ± ,\(\alpha=\pm \frac{\pi}{6}\),n∈Z

Question 14.
Find general solution of θ which satisfies both the equations sinθ = \(-\frac{1}{2}\) and cosθ = \(-\frac{\sqrt{3}}{2}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 12

Question 15.
Solve 4 sin x. sin 2x sin 4x = sin 3x
Solution:
Given sin 3x = 4 sinx sin 2x sin 4x
= 2sinx (2sin2x – sin4x)
⇒ 2 sin x (cos 2x – cos 6x)
⇒ sin3x = 2cos 2x sinx – 2 cos 6x sin x
⇒ sin 3x = sin 3x – sin x – 2 cos 6xsinx
⇒ 2cos 6x sinx + sinx= 0
⇒ sinx(2 cos 6x + 1)=0
⇒ sinx = 0 or cos 6x \(-\frac{1}{2}\)

Case (i): sin x = 0,  x= nπ, n∈Z is the general solution.
Case (ii) : cos 6x = \(-\frac{1}{2}\)
Principal solution is α \( =\frac{2 \pi}{3}\)
∴ General solution is 6x = 2πr ± \(\frac{2 \pi}{3}\)
\(x=\frac{n \pi}{3} \pm \frac{\pi}{9}, n \in Z\)

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 16.
If 0<θ<π solve cosθ cos2θ cos3θ = \(\frac{1}{4}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 12
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 13

Question 18.
Solve sin2x – cos2x = sinx – cosx
Solution:
The given equation can be written as
sin 2x – sin x = cos 2x – cos x
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 14

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 19.
Solve cos 3x + cos 2x = sin + sin = \(\sin \frac{3 x}{2}+\sin \frac{x}{2} 0 \leq x \leq 2 \pi\)
Solution:
Given cos 3x + cos 2x = sin + sin = \(\sin \frac{3 x}{2}+\sin \frac{x}{2} 0 \leq x \leq 2 \pi\)
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 15
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 16

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