Maths-
General
Easy

Question

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

  1. n pi plus-or-minus pi over 6 comma straight for all n element of Z
  2. n pi plus-or-minus pi over 4 comma straight for all n element of Z
  3. n pi plus-or-minus pi over 3 comma straight for all n element of Z
  4. n pi plus-or-minus pi over 2 comma straight for all n element of Z

Hint:

A general solution is one which involves the integer ‘n’ and gives all solutions of a trigonometric equation.

The correct answer is: n pi plus-or-minus pi over 6 comma straight for all n element of Z


    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).

    Sin2x + Cos2x = 1

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    Let the other three vectors be denoted as P with rightwards arrow on top comma stack Q space with rightwards arrow on top a n d space R with rightwards arrow on top
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    Q with rightwards arrow on top equals open parentheses b with rightwards arrow on top. i with hat on top close parentheses i with hat on top plus open parentheses b with rightwards arrow on top. j with hat on top close parentheses j with hat on top plus open parentheses b with rightwards arrow on top. k with hat on top close parentheses k with hat on top
    R with rightwards arrow on top equals i with hat on top plus j with hat on top minus 2 k with overparenthesis on top
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b with rightwards arrow on top. j with hat on top space equals left parenthesis i with hat on top minus j with hat on top right parenthesis left parenthesis j with hat on top right parenthesis space equals space minus 1
b with rightwards arrow on top. k with hat on top equals left parenthesis i with hat on top minus j with hat on top right parenthesis left parenthesis k with hat on top right parenthesis space equals space 0
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space space space space space space space space equals space 1 space minus space 1
space space space space space space space space space equals space 0
S o comma space t h e space g i v e n space v e c t o r s space a r e space p e r p e n d i c l u a r.
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space space space space space space space space space equals space 0
S o comma space t h e space g i v e n space v e c t o r s space a r e space p e r p e n d i c u l a r

P with rightwards arrow on top. R with rightwards arrow on top equals open parentheses i with hat on top plus j with hat on top plus k with hat on top close parentheses left parenthesis i with hat on top plus j with hat on top minus 2 k with hat on top right parenthesis
space space space space space space space space equals space 1 space plus space 1 space minus space 2
space space space space space space space space equals space 0

space space space space space space space space
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    If we see all the vectors are perpendicular to each other.
    The given vectors are mutually perpendicular.

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    Let the other three vectors be denoted as P with rightwards arrow on top comma stack Q space with rightwards arrow on top a n d space R with rightwards arrow on top
    P with rightwards arrow on top equals open parentheses a with rightwards arrow on top.1 with overparenthesis on top close parentheses i with overparenthesis on top plus open parentheses a with rightwards arrow on top. j with overparenthesis on top close parentheses i with overparenthesis on top plus open parentheses a with rightwards arrow on top. k with overparenthesis on top close parentheses k with hat on top
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    We have to find the relation between the vectors.
    Now, we will first find the components of the vectors
    a with rightwards arrow on top. i with overparenthesis on top equals left parenthesis i with hat on top plus i with hat on top plus k with hat on top right parenthesis open parentheses i with hat on top close parentheses equals 1
stack a. with rightwards arrow on top j with hat on top equals left parenthesis i with hat on top plus j with hat on top plus k with hat on top right parenthesis left parenthesis j with hat on top right parenthesis space equals space 1
a with rightwards arrow on top. k with hat on top equals left parenthesis i with hat on top plus j with hat on top plus k with hat on top right parenthesis k with hat on top equals 1
    b with rightwards arrow on top. i with overparenthesis on top equals open parentheses i with hat on top minus j with hat on top close parentheses open parentheses i with hat on top close parentheses equals 1
b with rightwards arrow on top. j with hat on top space equals left parenthesis i with hat on top minus j with hat on top right parenthesis left parenthesis j with hat on top right parenthesis space equals space minus 1
b with rightwards arrow on top. k with hat on top equals left parenthesis i with hat on top minus j with hat on top right parenthesis left parenthesis k with hat on top right parenthesis space equals space 0
    So, the vectors become.
    P with rightwards arrow on top equals i with hat on top plus j with hat on top plus k with hat on top
Q with rightwards arrow on top equals i with hat on top minus j with hat on top
    Now, we will take their dot product to check if they are mutually perpendicular or not.
    P with bar on top. Q with bar on top equals open parentheses i with hat on top plus j with hat on top plus k with hat on top close parentheses. open parentheses i with hat on top minus j with hat on top close parentheses
space space space space space space space space equals space 1 space minus space 1
space space space space space space space space space equals space 0
S o comma space t h e space g i v e n space v e c t o r s space a r e space p e r p e n d i c l u a r.
    Q with rightwards arrow on top. R with rightwards arrow on top equals open parentheses i with hat on top minus j with hat on top close parentheses open parentheses i with hat on top plus j with hat on top minus 2 k with hat on top close parentheses
space space space space space space space space equals space 1 minus 1
space space space space space space space space space equals space 0
S o comma space t h e space g i v e n space v e c t o r s space a r e space p e r p e n d i c u l a r

P with rightwards arrow on top. R with rightwards arrow on top equals open parentheses i with hat on top plus j with hat on top plus k with hat on top close parentheses left parenthesis i with hat on top plus j with hat on top minus 2 k with hat on top right parenthesis
space space space space space space space space equals space 1 space plus space 1 space minus space 2
space space space space space space space space equals space 0

space space space space space space space space
    So, the given vectors are perpendicular.
    If we see all the vectors are perpendicular to each other.
    The given vectors are mutually perpendicular.
    parallel

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