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Solution of Tan space x plus Tan space open parentheses 120 to the power of 6 plus x close parentheses plus Tan space open parentheses x minus 120 to the power of ring operator close parentheses equals 0 is

  1. fraction numerator n pi over denominator 3 end fraction comma straight for all n element of Z
  2. fraction numerator n pi over denominator 6 end fraction comma straight for all n element of Z
  3. fraction numerator n pi over denominator 4 end fraction comma straight for all n element of Z
  4. fraction numerator n pi over denominator 2 end fraction comma straight for all n element of Z

The correct answer is: fraction numerator n pi over denominator 3 end fraction comma straight for all n element of Z

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If Tan space bold A plus Tan space 2 bold A plus square root of 3 Tan space bold ATan 2 bold A equals square root of 3 then the general solution of A over 2 equals

If Tan space bold A plus Tan space 2 bold A plus square root of 3 Tan space bold ATan 2 bold A equals square root of 3 then the general solution of A over 2 equals

maths-General
General
physics-

The graph between the displacement x and t for a particle moving in a straight line is shown in figure. During the interval O A comma A B comma B C and C D comma the acceleration of the particle is

O A comma blank A B comma blank B C, C D

Region O A shows that graph bending toward time axis i. e. acceleration is negative.
Region A B shows that graph is parallel to time axisi. e. blankvelocity is zero. Hence acceleration is zero.
Region B C shows that graph is bending towards displacement axis i. e. acceleration is positive.
Region C D blankshows that graph having constant slope i. e. velocity is constant. Hence acceleration is zero

The graph between the displacement x and t for a particle moving in a straight line is shown in figure. During the interval O A comma A B comma B C and C D comma the acceleration of the particle is

O A comma blank A B comma blank B C, C D

physics-General
Region O A shows that graph bending toward time axis i. e. acceleration is negative.
Region A B shows that graph is parallel to time axisi. e. blankvelocity is zero. Hence acceleration is zero.
Region B C shows that graph is bending towards displacement axis i. e. acceleration is positive.
Region C D blankshows that graph having constant slope i. e. velocity is constant. Hence acceleration is zero
General
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tan space open parentheses pi over 4 plus x over 2 close parentheses minus tan space open parentheses pi over 4 minus x over 2 close parentheses equals 2 then the general solution of x =

tan space open parentheses pi over 4 plus x over 2 close parentheses minus tan space open parentheses pi over 4 minus x over 2 close parentheses equals 2 then the general solution of x =

maths-General
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physics-

Velocity-time left parenthesis v-t right parenthesis graph for a moving object is shown in the figure. Total displacement of the object during the same interval when there is non-zero acceleration and retardation is

Between time interval 20 blank s e c to 40 blank s e c, there is non-zero acceleration and retardation. Hence distance travelled during this interval
equals Area between time interval 20 blank s e c to 40 blank s e c
equals fraction numerator 1 over denominator 2 end fraction cross times 20 cross times 3 plus 20 cross times 1 equals 30 plus 20 equals 50 blank m

Velocity-time left parenthesis v-t right parenthesis graph for a moving object is shown in the figure. Total displacement of the object during the same interval when there is non-zero acceleration and retardation is

physics-General
Between time interval 20 blank s e c to 40 blank s e c, there is non-zero acceleration and retardation. Hence distance travelled during this interval
equals Area between time interval 20 blank s e c to 40 blank s e c
equals fraction numerator 1 over denominator 2 end fraction cross times 20 cross times 3 plus 20 cross times 1 equals 30 plus 20 equals 50 blank m
General
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In capital delta A B C comma R to the power of 2 end exponent left parenthesis blank s i n blank 2 A plus blank s i n blank 2 B plus blank s i n blank 2 C right parenthesis equals

Given R to the power of 2 end exponent left parenthesis blank s i n blank 2 A plus blank s i n blank 2 B plus blank s i n blank 2 C right parenthesis equals
R to the power of 2 end exponent( 4 blank s i n blankAsinBsin C)
equals 4 a to the power of 2 end exponent cross times fraction numerator a over denominator 2 a end fraction cross times fraction numerator b over denominator 2 a end fraction cross times fraction numerator c over denominator 2 a end fraction equals fraction numerator a b c over denominator 2 R end fraction equals 2 capital delta.

In capital delta A B C comma R to the power of 2 end exponent left parenthesis blank s i n blank 2 A plus blank s i n blank 2 B plus blank s i n blank 2 C right parenthesis equals

maths-General
Given R to the power of 2 end exponent left parenthesis blank s i n blank 2 A plus blank s i n blank 2 B plus blank s i n blank 2 C right parenthesis equals
R to the power of 2 end exponent( 4 blank s i n blankAsinBsin C)
equals 4 a to the power of 2 end exponent cross times fraction numerator a over denominator 2 a end fraction cross times fraction numerator b over denominator 2 a end fraction cross times fraction numerator c over denominator 2 a end fraction equals fraction numerator a b c over denominator 2 R end fraction equals 2 capital delta.
General
maths-

3 Sin space x plus 4 Cos space x minus 6 equals 0 then the general solution of bold x =

3 Sin space x plus 4 Cos space x minus 6 equals 0 then the general solution of bold x =

maths-General
General
maths-

Solution of Cot squared space theta plus open parentheses square root of 3 plus fraction numerator 1 over denominator square root of 3 end fraction close parentheses Cot space theta plus 1 equals 0 is

Solution of Cot squared space theta plus open parentheses square root of 3 plus fraction numerator 1 over denominator square root of 3 end fraction close parentheses Cot space theta plus 1 equals 0 is

maths-General
General
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The general solution of  fraction numerator tan space 5 x minus tan space 4 x over denominator 1 plus tan space 5 x tan space 4 x end fraction equals 1 is

The general solution of  fraction numerator tan space 5 x minus tan space 4 x over denominator 1 plus tan space 5 x tan space 4 x end fraction equals 1 is

maths-General
General
maths-

If Sin space A equals Sin space B comma Cos space A equals Cos space B text  then  end text A equals

If Sin space A equals Sin space B comma Cos space A equals Cos space B text  then  end text A equals

maths-General
General
maths-

In capital delta A B C comma capital sigma a to the power of 3 end exponent blank c o s blank(B-C) equals

A equals B equals C equals 6 0 to the power of 0 end exponent comma a equals b equals c equals 1

In capital delta A B C comma capital sigma a to the power of 3 end exponent blank c o s blank(B-C) equals

maths-General
A equals B equals C equals 6 0 to the power of 0 end exponent comma a equals b equals c equals 1
General
maths-

If cot space theta minus tan space theta equals 2 then principal value of theta

If cot space theta minus tan space theta equals 2 then principal value of theta

maths-General
General
maths-

If 3cos θsin θ=1 then θ=

If 3cos θsin θ=1 then θ=

maths-General
General
maths-

If straight A and B are acute angles such that  Sin space A equals Sin squared space B and 2 Cos squared space A equals 3 Cos squared space B then A equals

If straight A and B are acute angles such that  Sin space A equals Sin squared space B and 2 Cos squared space A equals 3 Cos squared space B then A equals

maths-General
General
physics-

Three persons P comma Q and R of same mass travel with same speed u such that each one faces the other always. After how much time will they meet each other?

Three persons P comma Q and R of same mass travel with same speed u such that each one faces the other always. After how much time will they meet each other?

physics-General
General
physics-

Figure given shows the distance–time graph of the motion of a car. It follows from the graph that the car is

Since x equals 1.2 t to the power of 2 end exponent which is in form x equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent
Thus, the motion is uniformly accelerated

Figure given shows the distance–time graph of the motion of a car. It follows from the graph that the car is

physics-General
Since x equals 1.2 t to the power of 2 end exponent which is in form x equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent
Thus, the motion is uniformly accelerated