मराठी

If 2sin2θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.

Advertisements
Advertisements

प्रश्न

If 2sin2θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.

बेरीज
Advertisements

उत्तर

2sin2θ = 3cosθ

We know that,

sin2θ = 1 – cos2θ

Given that,

2sin2θ = 3 cosθ

2 – 2cos2θ = 3cosθ

2cos2θ + 3cosθ – 2 = 0

(cosθ + 2)(2cosθ – 1) = 0

Therefore,

cosθ = `1/2 = cos  pi/3`

θ = `pi/3` or `2π  –  pi/3`

θ = `pi/3, (5pi)/3`

Therefore, 2(1 – cos2θ) = 3cosθ

⇒ 2 – 2cos2θ = 3cosθ

⇒ 2cos2θ + 3cosθ – 2 = 0

⇒ 2cos2θ + 4cosθ – cosθ – 2 = 0

⇒ 2cosθ(cosθ + 2) + 1(cosθ + 2) = 0

⇒ (2cosθ + 1)(cosθ + 2) = 0

Since, cosθ ∈ [–1, 1], for any value θ.

So, cosθ ≠ –2

Therefore,

2cosθ – 1 = 0

⇒ cosθ = `1/2`

= `pi/3` or `2π  –  pi/3`

θ = `π/3, (5pi)/3`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometric Functions - Exercise [पृष्ठ ५४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 3 Trigonometric Functions
Exercise | Q 18 | पृष्ठ ५४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

If \[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\], then the values of tan x, sec x and cosec x


Prove that: cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = \[\frac{1}{2}\]


Prove that

\[\frac{cosec(90^\circ + x) + \cot(450^\circ + x)}{cosec(90^\circ - x) + \tan(180^\circ - x)} + \frac{\tan(180^\circ + x) + \sec(180^\circ - x)}{\tan(360^\circ + x) - \sec( - x)} = 2\]

 


Prove that

\[\frac{\sin(180^\circ + x) \cos(90^\circ + x) \tan(270^\circ - x) \cot(360^\circ - x)}{\sin(360^\circ - x) \cos(360^\circ + x) cosec( - x) \sin(270^\circ + x)} = 1\]

 


Prove that:
\[\sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2\]

 

Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]


Prove that:
\[\sin \frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac{1}{2}\]


If \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] is equal to

 


If tan A + cot A = 4, then tan4 A + cot4 A is equal to


If sec x + tan x = k, cos x =


Which of the following is incorrect?


Find the general solution of the following equation:

\[\sin x = \frac{1}{2}\]

Find the general solution of the following equation:

\[\tan x + \cot 2x = 0\]

Find the general solution of the following equation:

\[\sin 3x + \cos 2x = 0\]

Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]


Solve the following equation:

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\]

Solve the following equation:

\[\sin 3x - \sin x = 4 \cos^2 x - 2\]

Solve the following equation:
\[\sec x\cos5x + 1 = 0, 0 < x < \frac{\pi}{2}\]


Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]


If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.


Write the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π].


Write the number of solutions of the equation
\[4 \sin x - 3 \cos x = 7\]


If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.

 

Write the number of points of intersection of the curves

\[2y = 1\] and \[y = \cos x, 0 \leq x \leq 2\pi\].
 

The general solution of the equation \[7 \cos^2 x + 3 \sin^2 x = 4\] is


A solution of the equation \[\cos^2 x + \sin x + 1 = 0\], lies in the interval


If \[4 \sin^2 x = 1\], then the values of x are

 


The equation \[3 \cos x + 4 \sin x = 6\] has .... solution.


Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

sin4x = sin2x


Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

2 sin2x + 1 = 3 sin x


Solve the following equations:
sin 5x − sin x = cos 3


Solve the following equations:
sin θ + sin 3θ + sin 5θ = 0


Solve the following equations:
2cos 2x – 7 cos x + 3 = 0


Choose the correct alternative:
If sin α + cos α = b, then sin 2α is equal to


Solve the equation sin θ + sin 3θ + sin 5θ = 0


Solve `sqrt(3)` cos θ + sin θ = `sqrt(2)`


If a cosθ + b sinθ = m and a sinθ - b cosθ = n, then show that a2 + b2 = m2 + n2 


Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×