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Principal Solution of Trigonometric Equation

Solution set ...

Question

Solution set of the equation ${sin}^{2} x + {cos}^{2} 3 x$ =1$ is given by

A

{\frac{n π}{4}, n ϵ I}

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B

{n π, n ϵ I}

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C

{\frac{n π}{2}, n ϵ I}

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D

n o n e o f t h e s e

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Solution

The correct option is B ${n π, n ϵ I}$
According to question,

${sin}^{2} x + {cos}^{2} 3 x = 1$

$⟹ c o s^{2} 3 x = 1 - {sin}^{2} x$
$⟹ {cos}^{2} 3 x = {cos}^{2} x$

Solving through trigonometric eq,

$3 x = 2 n π + x$
$⟹ 2 x = 2 n π$

=> $x = n π$ , $n ϵ I$

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Similar questions

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be four non-zero numbers satisfying the equation $t a n^{- 1} (\frac{a}{x}) + t a n^{- 1} (\frac{b}{x}) + t a n^{- 1} (\frac{c}{x}) + t a n^{- 1} (\frac{d}{x}) = \frac{π}{2},$ then which of the following relation hold(s) good?

Q. The number of real solutions of the equation

{sin}^{- 1} (\infty \sum i = 1 x^{i + 1} - x \infty \sum i = 1 {(\frac{x}{2})}^{i}) = \frac{π}{2} - {cos}^{- 1} (\infty \sum i = 1 {(- \frac{x}{2})}^{i} - \infty \sum i = 1 (- x)^{i})

lying in the interval

(- \frac{1}{2}, \frac{1}{2})

is .
(Here, the inverse trigonometric functions

{sin}^{- 1} x

{cos}^{- 1} x

assume values in

[- \frac{π}{2}, \frac{π}{2}]

[0, π]

, respectively.)

Q. Complex numbers which satisfy both the equations

| z - 1 - i | = \sqrt{2}

| z + 1 + i | = 2

Q. The number of real solutions of the equation

{sin}^{- 1} (\infty \sum i = 1 x^{i + 1} - x \infty \sum i = 1 {(\frac{x}{2})}^{i}) = \frac{π}{2} - {cos}^{- 1} (\infty \sum i = 1 {(- \frac{x}{2})}^{i} - \infty \sum i = 1 (- x)^{i})

lying in the interval

(- \frac{1}{2}, \frac{1}{2})

Q. Assertion :If

f (x) = (cos x + i sin x) (cos 2 x + i sin 2 x)

(cos 3 x + i sin 3 x) . . . (cos n x + i sin n x)

f (1) = 1

f^{''} (1)

- {(\frac{n (n + 1)}{2})}^{2}

f (x) = cos \frac{n (n - 1)}{2} x + i sin \frac{n (n - 1)}{2} x

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