मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Solve the following equations:cos 2θ = 5+14 - Mathematics

Advertisements
Advertisements

प्रश्न

Solve the following equations:
cos 2θ = `(sqrt(5) + 1)/4`

बेरीज

उत्तर

We know cos 36° = `(sqrt(5) + 1)/4`, 36° = `pi/5`

cos 2θ = cos 36° = `cos (pi/5)`

The general solution is

2θ = `2"n"pi +- pi/5`, n ∈ Z

 θ = `"n"pi +- pi/10`, n ∈ Z

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - Exercise 3.8 [पृष्ठ १३३]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.8 | Q 3. (x) | पृष्ठ १३३

संबंधित प्रश्‍न

Find the principal and general solutions of the equation sec x = 2


Find the general solution of the equation sin 2x + cos x = 0


If \[\tan x = \frac{b}{a}\] , then find the values of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\].


If \[cosec x - \sin x = a^3 , \sec x - \cos x = b^3\], then prove that \[a^2 b^2 \left( a^2 + b^2 \right) = 1\]


Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]


Prove that:  tan 225° cot 405° + tan 765° cot 675° = 0


Prove that: \[\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4} {cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3 - 4\sqrt{3}}{2}\]

 


In a ∆ABC, prove that:

\[\tan\frac{A + B}{2} = \cot\frac{C}{2}\]

If x = r sin θ cos ϕ, y = r sin θ sin ϕ and r cos θ, then x2 + y2 + z2 is independent of


If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]

 

The value of sin25° + sin210° + sin215° + ... + sin285° + sin290° is


sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 =


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


Find the general solution of the following equation:

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

Solve the following equation:

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

Solve the following equation:

\[\tan 3x + \tan x = 2\tan 2x\]

Solve the following equation:
\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]


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

 


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


Solve the following equations:
cot θ + cosec θ = `sqrt(3)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×