Maths-
General
Easy

Question

The general value of theta satisfies the equation Tan space theta Tan space open parentheses 120 to the power of ring operator plus theta close parentheses Tan space open parentheses 120 to the power of ring operator minus theta close parentheses equals fraction numerator 1 over denominator square root of 3 end fraction is

  1. open left parenthesis 6 n plus 1 right parenthesis pi over 18 comma straight for all n element of Z 2 close parentheses
  2. open left parenthesis 3 n plus 1 right parenthesis pi over 3 comma straight for all n element of Z close parentheses
  3. open left parenthesis 6 n plus 1 right parenthesis pi over 6 comma straight for all n element of Z space 4 close parentheses
  4. open left parenthesis 3 n plus 1 right parenthesis pi over 6 comma straight for all n element of Z close parentheses

The correct answer is: open left parenthesis 6 n plus 1 right parenthesis pi over 18 comma straight for all n element of Z 2 close parentheses

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A block of weight W is suspended by a string of fixed length. The ends of the string are held at various positions as shown in the figures below. In which case, if any, is the magnitude of the tension along the string largest?

A block of weight W is suspended by a string of fixed length. The ends of the string are held at various positions as shown in the figures below. In which case, if any, is the magnitude of the tension along the string largest?

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If Tan space x plus 2 Tan space 2 x plus 4 Tan space 4 x plus 8 C cot space 8 x equals square root of 3 then the general solution of x=

If Tan space x plus 2 Tan space 2 x plus 4 Tan space 4 x plus 8 C cot space 8 x equals square root of 3 then the general solution of x=

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An ideal string is passing over a smooth pulley as shown. Two blocks  and  are connected at the ends of the string. If  = 1 kg and tension in the string is 10 N, mass m2 is equal to open parentheses straight g equals 10 straight m divided by straight s squared close parentheses

An ideal string is passing over a smooth pulley as shown. Two blocks  and  are connected at the ends of the string. If  = 1 kg and tension in the string is 10 N, mass m2 is equal to open parentheses straight g equals 10 straight m divided by straight s squared close parentheses

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Tan space open parentheses pi over 4 plus A over 2 close parentheses plus Tan space open parentheses pi over 4 minus A over 2 close parentheses equals fraction numerator 4 over denominator square root of 3 end fraction; then the general solution of A=

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Three forces ,  and  act on an object simultaneously. These force vectors are shown in the following free-body diagram. In which direction does the object accelerate?

Three forces ,  and  act on an object simultaneously. These force vectors are shown in the following free-body diagram. In which direction does the object accelerate?

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The general solution of sin space 2 x equals 4 cos space x is

The general solution of sin space 2 x equals 4 cos space x is

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If 11 sin squared space x plus 7 Cos squared space x equals 8 then x = - ....

Here, we have to find the value of x.
11 sin2x + 7 cos2x = 8 .
Or, 11 sin2x + 7 ( 1 - sin2x) = 8 .
Or, 11 sin2x + 7 - 7sin2x = 8 .
Or, 11sin2x - 7 sin2x = 8 - 7 .
Or, 4sin2x = 1
Or, sin2x = 1/4
Or, Sin2 x= Sinπ/6
x = π/6
Hence, the correct option is (c).

If 11 sin squared space x plus 7 Cos squared space x equals 8 then x = - ....

Maths-General
Here, we have to find the value of x.
11 sin2x + 7 cos2x = 8 .
Or, 11 sin2x + 7 ( 1 - sin2x) = 8 .
Or, 11 sin2x + 7 - 7sin2x = 8 .
Or, 11sin2x - 7 sin2x = 8 - 7 .
Or, 4sin2x = 1
Or, sin2x = 1/4
Or, Sin2 x= Sinπ/6
x = π/6
Hence, the correct option is (c).
General
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The general solution of fraction numerator Tan space 2 x minus Tan space x over denominator 1 plus Tan space x times Tan space 2 x end fraction equals 1 i

Here, we have to find the general solution.
Tan 2x - Tan x / 1 + Tan x. Tan 2x = 1 .
Or, Tan (2x - 1 ) = 1 .
Or, Tan x = 1 .
Hence, the correct option is (d).

The general solution of fraction numerator Tan space 2 x minus Tan space x over denominator 1 plus Tan space x times Tan space 2 x end fraction equals 1 i

Maths-General
Here, we have to find the general solution.
Tan 2x - Tan x / 1 + Tan x. Tan 2x = 1 .
Or, Tan (2x - 1 ) = 1 .
Or, Tan x = 1 .
Hence, the correct option is (d).
General
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If Error converting from MathML to accessible text. then Error converting from MathML to accessible text.

If Error converting from MathML to accessible text.
then Error converting from MathML to accessible text. = 0

If Error converting from MathML to accessible text. then Error converting from MathML to accessible text.

Maths-General
If Error converting from MathML to accessible text.
then Error converting from MathML to accessible text. = 0
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The most general value of theta satisfying the equations Sin space theta equals fraction numerator 1 over denominator square root of 2 end fraction comma Cos space theta equals negative fraction numerator 1 over denominator square root of 2 end fraction is

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The smallest value of theta satisfying the equation square root of 3 left parenthesis Cot space theta plus Tan space theta right parenthesis equals 4 IS

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maths-General
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maths-

If sin space 2 theta minus cos space 3 theta equals 0 and theta is acute then theta

If sin space 2 theta minus cos space 3 theta equals 0 and theta is acute then theta

maths-General
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maths-

If sin space open parentheses x plus 28 to the power of ring operator close parentheses equals cos space open parentheses 3 x minus 78 to the power of ring operator close parentheses then x=

If sin space open parentheses x plus 28 to the power of ring operator close parentheses equals cos space open parentheses 3 x minus 78 to the power of ring operator close parentheses then x=

maths-General
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The adjoining figure shows a force of 40 N pulling a body of mass 5 kg in a direction  above the horizontal. The body is in rest on a smooth horizontal surface. Assuming acceleration of free-fall is 10 . Which of the following statements I and II is/are correct?

I) The weight of the 5 kg mass acts vertically downwards

II) The net vertical force acting on the body is 30 N.

The adjoining figure shows a force of 40 N pulling a body of mass 5 kg in a direction  above the horizontal. The body is in rest on a smooth horizontal surface. Assuming acceleration of free-fall is 10 . Which of the following statements I and II is/are correct?

I) The weight of the 5 kg mass acts vertically downwards

II) The net vertical force acting on the body is 30 N.

physics-General
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A sphere of mass m is kept in equilibrium with the help of several springs as shown in the figure. Measurement shows that one of the springs applies a force   on the sphere. With what acceleration the sphere will move immediately after this particular spring is cut?

A sphere of mass m is kept in equilibrium with the help of several springs as shown in the figure. Measurement shows that one of the springs applies a force   on the sphere. With what acceleration the sphere will move immediately after this particular spring is cut?

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