JEE Main Previous Year Trigonometry Questions and Solutions - Solution of Triangles Trigonometric equations

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Suppose $θ \in [0, \frac{π}{4}]$ is a solution of $4 \cos θ - 3 \sin θ = 1$ . Then $\cos θ$ is equal to : [2024]

1 $\frac{6 - \sqrt{6}}{(3 \sqrt{6} - 2)}$

2 $\frac{4}{(3 \sqrt{6} - 2)}$

3 $\frac{4}{(3 \sqrt{6} + 2)}$

4 $\frac{6 + \sqrt{6}}{(3 \sqrt{6} + 2)}$

Ans.
(2)

We have, $4 \cos θ - 3 \sin θ = 1$

$\Rightarrow 4 \cos θ - 1 = 3 \sin θ \Rightarrow {(4 \cos θ - 1)}^{2} = {(3 \sin θ)}^{2}$

$\Rightarrow 16 \cos^{2} θ + 1 - 8 \cos θ = 9 \sin^{2} θ$

$\Rightarrow 16 \cos^{2} θ + 1 - 8 \cos θ = 9 (1 - \cos^{2} θ)$

$\Rightarrow 25 \cos^{2} θ - 8 \cos θ - 8 = 0$

$\Rightarrow \cos θ = \frac{8 \pm \sqrt{64 + 800}}{50}$

$= \frac{8 \pm \sqrt{864}}{50} = \frac{8 \pm 12 \sqrt{6}}{50} = \frac{4 \pm 6 \sqrt{6}}{25}$

Since, $θ \in [0, π / 4]$ so $\cos θ = \frac{4 + 6 \sqrt{6}}{25}$

$= \frac{(4 - 6 \sqrt{6}) (4 + 6 \sqrt{6})}{25 (4 - 6 \sqrt{6})} = \frac{- 200}{25 \times 2 (2 - 3 \sqrt{6})} = \frac{4}{3 \sqrt{6} - 2}$

Want Us to Email you About Special Offers & Updates?