Given-sin-5x-38-cos-2x-16-0-x-90-find-the-value-of-x-

Question Number 161060 by pete last updated on 11/Dec/21

Given \sin (5 x - 38) = \cos (2 x + 16), 0 ° ⩽ x ⩽ 90 °,

find the value of x

Commented by cortano last updated on 11/Dec/21

r e c a l l \sin (90 ° - x) = \cos x

\Leftrightarrow \sin (5 x - 38 °) = \sin (90 ° - (128 ° - 5 x)) = \cos (128 ° - 5 x)

\Leftrightarrow \cos (128 ° - 5 x) = \cos (2 x + 16 °)

\Leftrightarrow 2 x + 16 ° = \pm (128 ° - 5 x) + 2 n π

\Leftrightarrow {\begin{cases} 2 x + 16 ° = 128 ° - 5 x + 2 n π \\ 2 x + 16 ° = 5 x - 128 ° + 2 n π \end{cases}

\Leftrightarrow {\begin{cases} 7 x = 112 ° + 2 n π \\ 3 x = 144 ° + 2 k π \end{cases}

\Leftrightarrow {\begin{cases} x = 16 ° + \frac{2 n π}{7} \\ x = 48 ° + \frac{2 k π}{3} \end{cases}

Commented by pete last updated on 11/Dec/21

thanks very much sir

Answered by mr W last updated on 11/Dec/21

\sin (5 x - 38) = \cos (2 x + 16)

\cos (90 - 5 x + 38) = \cos (2 x + 16)

90 - 5 x + 38 = 360 k \pm (2 x + 16)

90 - 5 x + 38 = 360 k + 2 x + 16

112 = 360 k + 7 x

\Rightarrow x = \frac{360 k}{7} + 16 = 16 °, 67 \frac{3}{7} °

90 - 5 x + 38 = 360 k - 2 x - 16

144 = 360 k + 3 x

\Rightarrow x = 120 k + 48 = 48 °

Commented by pete last updated on 11/Dec/21

thanks sir

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