$If space sin open parentheses sin to the power of negative 1 end exponent 1 fifth plus cos to the power of negative 1 end exponent straight x close parentheses equals 1$ then find the value of x.

Solution

The given equation is
$space sin open parentheses sin to the power of negative 1 end exponent 1 fifth plus cos to the power of negative 1 end exponent straight x close parentheses equals 1 space space space space space space space space space space space space or space space sin to the power of negative 1 end exponent 1 fifth plus cos to the power of negative 1 end exponent straight x equals straight pi over 2 or space space sin to the power of negative 1 end exponent 1 fifth equals straight pi over 2 minus cos to the power of negative 1 end exponent straight x space space space space space space space space space space space space or space sin to the power of negative 1 end exponent 1 fifth equals sin to the power of negative 1 end exponent straight x rightwards double arrow space space space space space space space space straight x equals 1 fifth$

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Inverse Trigonometric Functions

$If space sin open parentheses sin to the power of negative 1 end exponent 1 fifth plus cos to the power of negative 1 end exponent straight x close parentheses equals 1$ then find the value of x.

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