How do you solve for X in cos(x+30)=sin(5x+12)?

Question

I dont get it, like i get trig and other stuff, but i dont know how to begin solving this problem

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Answer 1 · 2017-01-24T18:09:51

Please see the explanation.

There are two base angles that solve this, $8^{\circ} and 27^{\circ}$ $8^@ and 27^@$ .

The first angle is found by converting the cosine function into the negative of the sine function by subtracting $90^{\circ}$ $90^@$ :

$- sin (x - 60^{\circ}) = sin (5 x + 12^{\circ})$ $-sin(x - 60^@) = sin(5x + 12^@)$

Use the identity $- sin (a) = sin (- a)$ $-sin(a) = sin(-a)$

$sin (- x + 60^{\circ}) = sin (5 x + 12^{\circ})$ $sin(-x + 60^@) = sin(5x + 12^@)$

Equate the arguments:

$- x + 60^{\circ} = 5 x + 12^{\circ}$ $-x + 60^@ = 5x + 12^@$

$6 x = 48^{\circ}$ $6x = 48^@$

$x = 8^{\circ}$ $x = 8^@$

Because $5 x - - x = 6 x$ $5x - -x = 6x$ , this repeats 6 times every cycle or every $60^{\circ}$ $60^@$

$x = 8^{\circ} + n 60^{\circ}$ $x = 8^@ + n60^@$ where n is any integer positive or negative including zero.

The second angle is found by converting the sine function into the cosine function by subtracting $90^{\circ}$ $90^@$ :

$cos (x + 30^{\circ}) = cos (5 x - 78^{\circ})$ $cos(x + 30^@) = cos(5x - 78^@)$

Equate the arguments:

$x + 30^{\circ} = 5 x - 78^{\circ}$ $x + 30^@ = 5x - 78^@$

$4 x = 128^{\circ}$ $4x = 128^@$

$x = 27^{\circ}$ $x = 27^@$

Because $5 x - x = 4 x$ $5x - x = 4x$ this repeats 4 times every cycle or every $90^{\circ}$ $90^@$

$x = 27^{\circ} + n 90^{\circ}$ $x = 27^@ + n90^@$

The complete answer that will give you all of the solutions is:

$x ={(8^@ + n60^@),(27^@+n90^@):}$ where n is any positive or negative integer, including 0.