In this question, we have to find the number of solution. In each quadrant tan value is different. Make two different case, one is where tan is positive ,[ 0 , π/2 ] U [π , 3 π/2 ] . And second case , tan is negative at [π/2 , π] U [3 π/2 , 2 π ] .

$vertical line tan space x vertical line equals tan space x plus fraction numerator 1 over denominator cos begin display style space end style x end fraction left parenthesis 0 less or equal than x less or equal than 2 pi right parenthesis$ has

Maths-General

General

maths-

If $A equals sin to the power of 2 end exponent invisible function application theta plus cos to the power of 4 end exponent invisible function application theta comma$ then for all real values of q

maths-General

General

chemistry-

The products (A) , (B) and (C) are:

chemistry-General

General

maths-

In the interval $left square bracket negative pi divided by 4 comma pi divided by 2 right square bracket$ , the equation, $cos space 4 x plus fraction numerator 10 tan space x over denominator 1 plus tan squared space x end fraction equals 3$ has

maths-General

General

chemistry-

$Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript plus straight X not stretchy ⟶ with straight H to the power of ⊖ on top Cr to the power of 3 plus end exponent plus straight H subscript 2 straight O plus$ oxidised product of , in the above reaction cannot be

chemistry-General

General

maths-

Let a, b, c, d $element of$ R. Then the cubic equation of the type $a x cubed plus b x squared plus c x plus d equals 0$ has either one root real or all three roots are real. But in case of trigonometric equations of the type $sin cubed space x plus b sin squared space x plus c sin space x plus d equals 0$ can possess several solutions depending upon the domain of x. To solve an equation of the type a $cos space theta plus b sin space theta equals c$ . The equation can be written as $cos space left parenthesis theta minus alpha right parenthesis equals c divided by square root of open parentheses a squared plus b squared close parentheses end root$ The solution is $theta equals 2 straight n pi plus alpha plus-or-minus beta$ where $tan space alpha$ = $b divided by a comma cos space beta equals c divided by square root of open parentheses a squared plus b squared close parentheses end root$
On the domain [– $straight pi$ , $straight pi$ ] the equation $4 sin cubed space x plus 2 sin squared space x minus 2 sin space x minus 1 equals 0$ possess

maths-General

General

maths-

$cos to the power of 4 invisible function application pi over 8 plus cos to the power of 4 invisible function application fraction numerator 3 pi over denominator 8 end fraction plus cos to the power of 4 invisible function application fraction numerator 5 pi over denominator 8 end fraction plus cos to the power of 4 invisible function application fraction numerator 7 pi over denominator 8 end fraction equals$

maths-General

General

chemistry-

The products (A) and (B) are:

chemistry-General

General

maths-

If w is a complex cube root of unity, then the matrix A = $open square brackets table row 1 cell w to the power of 2 end exponent end cell w row cell w to the power of 2 end exponent end cell w 1 row w 1 cell w to the power of 2 end exponent end cell end table close square brackets$ is a-

maths-General

General

maths-

Matrix [1 2] $open parentheses open square brackets table row cell negative 2 end cell 5 row 3 2 end table close square brackets open square brackets table row 1 row 2 end table close square brackets close parentheses$ is equal to-

maths-General

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The oxidation number of carboxylic carbon atom in is:

+2+4+1+3

The correct answer is: +3

Related Questions to study

The oxidation number of carboxylic carbon atom in is:

The oxidation number of carboxylic carbon atom in is:

The product (A) is:

The product (A) is:

If then is always

If then is always

The compound (A) is:

The compound (A) is:

The oxidation number of C in is:

The oxidation number of C in is:

has

has

If then for all real values of q

If then for all real values of q

The products (A) , (B) and (C) are:

The products (A) , (B) and (C) are:

In the interval , the equation, has

In the interval , the equation, has

oxidised product of , in the above reaction cannot be

oxidised product of , in the above reaction cannot be

The products (A) and (B) are:

The products (A) and (B) are:

If w is a complex cube root of unity, then the matrix A = is a-

If w is a complex cube root of unity, then the matrix A = is a-

Matrix [1 2] is equal to-

Matrix [1 2] is equal to-

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The oxidation number of carboxylic carbon atom in $CH subscript 3 COOH$ is:

+2
+4
+1
+3

The oxidation number of carboxylic carbon atom in $CH subscript 3 COOH$ is:

The oxidation number of carboxylic carbon atom in $CH subscript 3 COOH$ is:

If $A plus B plus C equals pi comma$ then $tan squared invisible function application A over 2 plus tan squared invisible function application B over 2 plus tan squared invisible function application C over 2$ is always

If $A plus B plus C equals pi comma$ then $tan squared invisible function application A over 2 plus tan squared invisible function application B over 2 plus tan squared invisible function application C over 2$ is always

The oxidation number of C in $CH subscript 2 straight O$ is:

The oxidation number of C in $CH subscript 2 straight O$ is:

$vertical line tan space x vertical line equals tan space x plus fraction numerator 1 over denominator cos begin display style space end style x end fraction left parenthesis 0 less or equal than x less or equal than 2 pi right parenthesis$ has

$vertical line tan space x vertical line equals tan space x plus fraction numerator 1 over denominator cos begin display style space end style x end fraction left parenthesis 0 less or equal than x less or equal than 2 pi right parenthesis$ has

If $A equals sin to the power of 2 end exponent invisible function application theta plus cos to the power of 4 end exponent invisible function application theta comma$ then for all real values of q

If $A equals sin to the power of 2 end exponent invisible function application theta plus cos to the power of 4 end exponent invisible function application theta comma$ then for all real values of q

In the interval $left square bracket negative pi divided by 4 comma pi divided by 2 right square bracket$ , the equation, $cos space 4 x plus fraction numerator 10 tan space x over denominator 1 plus tan squared space x end fraction equals 3$ has

In the interval $left square bracket negative pi divided by 4 comma pi divided by 2 right square bracket$ , the equation, $cos space 4 x plus fraction numerator 10 tan space x over denominator 1 plus tan squared space x end fraction equals 3$ has

$Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript plus straight X not stretchy ⟶ with straight H to the power of ⊖ on top Cr to the power of 3 plus end exponent plus straight H subscript 2 straight O plus$ oxidised product of , in the above reaction cannot be

$Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript plus straight X not stretchy ⟶ with straight H to the power of ⊖ on top Cr to the power of 3 plus end exponent plus straight H subscript 2 straight O plus$ oxidised product of , in the above reaction cannot be

If w is a complex cube root of unity, then the matrix A = $open square brackets table row 1 cell w to the power of 2 end exponent end cell w row cell w to the power of 2 end exponent end cell w 1 row w 1 cell w to the power of 2 end exponent end cell end table close square brackets$ is a-

If w is a complex cube root of unity, then the matrix A = $open square brackets table row 1 cell w to the power of 2 end exponent end cell w row cell w to the power of 2 end exponent end cell w 1 row w 1 cell w to the power of 2 end exponent end cell end table close square brackets$ is a-

Matrix [1 2] $open parentheses open square brackets table row cell negative 2 end cell 5 row 3 2 end table close square brackets open square brackets table row 1 row 2 end table close square brackets close parentheses$ is equal to-

Matrix [1 2] $open parentheses open square brackets table row cell negative 2 end cell 5 row 3 2 end table close square brackets open square brackets table row 1 row 2 end table close square brackets close parentheses$ is equal to-