Solve the following trigonometric equation where 0 ∘ < q < 90 ∘ . 2 cos 2 q - 3 cos q = 0

Given: 2 cos 2 q - 3 cos q = 0 2 cos 2 q - 3 cos q = 0 2 cos 2 q = 3 cos q cos 2 q cos q = 3 2 cos q = 3 2 ∴ q = 30 ∘ Hence, the solution for the given equation is q = 30 ∘ .

Solve the following trigonometric equation where 0 ∘ < q < 90 ∘ . 2 sin 2 q - cos q = 1

Given: 2 sin 2 q - cos q = 1 2 sin 2 q - cos q = 1 2 1 - cos 2 q - cos q = 1 2 - 2 cos 2 q - cos q - 1 = 0 1 - 2 cos 2 q - cos q = 0 2 cos 2 q + cos - 1 = 0 2 cos 2 q + 2 cos q - cos q - 1 = 0 2 cos q cos q + 1 - cos q + 1 = 0 cos q + 1 2 cos q - 1 = 0 cos q = - 1 and cosq = 1 2 ∴ q = 60 ° Since cos q cannot be negative where 0 ° < q < 90 ° , therefore we will ignore cos q = - 1 . Hence, the solution of the given equation is q = 60 °

Solve the following trigonometric equation where 0 ∘ < q < 90 ∘ . 3 tan 2 q = 2 s e c 2 q + 1

Given: 3 tan 2 q = 2 s e c 2 q + 1 3 tan 2 q = 2 s e c 2 q + 1 3 tan 2 q = 2 tan 2 q + 1 + 1 3 tan 2 q = 2 tan 2 q + 2 + 1 3 tan 2 q - 2 tan 2 q = 3 tan 2 q = 3 tan q = 3 q = 30 ∘ Hence, the solution for the given equation is q = 30 ∘ .

Solve the following trigonometric equation where 0 ∘ < q < 90 ∘ . cos 2 q - 3 cos q + 2 = sin 2 q

Given: cos 2 q - 3 cos q + 2 = sin 2 q cos 2 q - 3 cos q + 2 = sin 2 q cos 2 q - 3 cos q + 2 = 1 - cos 2 q cos 2 q - 3 cos q + cos 2 q = 1 - 2 2 cos 2 q - 3 cos q + 1 = 0 2 cos 2 q - 2 cos q - 1 cos q + 1 = 0 2 cos q cos q - 1 - cos q - 1 = 0 cos q - 1 2 cos q - 1 = 0 cos q = 1 and cos q = 1 2 ∴ q = 0 ∘ and 60 ∘ As 0 ° < q < 90 ° . Hence, solution of the given equation is q = 60 ∘ .

Solve the following trigonometric equation where 0 ∘ < q < 90 ∘ . cos q 1 - sin q + cos q 1 + sin q = 4

Given: cos q 1 - sin q + cos q 1 + sin q = 4 cos q 1 - sin q + cos q 1 + sin q = 4 cos q 1 + sin q + cos q 1 - sin q 1 - sin q 1 + sin q = 4 cos q + sin q · cos q + cos q - cos q · sin q 1 - sin 2 q = 4 2 cos q cos 2 q = 4 ∵ sin 2 q + cos 2 q = 1 cos q = 1 2 q = 60 ∘ Hence, the solution of given equation is q = 60 ∘ .

E M B I B E

Mathematics Class - 10>Trigonometry>Chapter 9 - Trigonometric Equation and Identities>Exercise – 2

Chhattisgarh Board Solutions for Exercise 6: Exercise – 2

Author:Chhattisgarh Board

Chhattisgarh Board Mathematics Solutions for Exercise - Chhattisgarh Board Solutions for Exercise 6: Exercise – 2

Attempt the free practice questions from Exercise 6: Exercise – 2 with hints and solutions to strengthen your understanding. Mathematics Class - 10 solutions are prepared by Experienced Embibe Experts.

Questions from Chhattisgarh Board Solutions for Exercise 6: Exercise – 2 with Hints & Solutions

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 9 - Trigonometric Equation and Identities>Try These>Q 2

Put $q = 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}$ in the given relations and check which value of $q$ satisfies the following relation.

$\tan^{2} q + c o t^{2} q = 2$

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 9 - Trigonometric Equation and Identities>Try These>Q 3

Put $q = 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}$ in the given relations and check which value of $q$ satisfies these relations.

$q = 60 °$

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 9 - Trigonometric Equation and Identities>Try These>Q 4

Put $q = 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}$ in the given relations and check which value of $q$ satisfies the following relation.