Answer:
x = π/12, 5π/12
Step-by-step explanation:
We can work this problem a couple of ways. We can convert it to single trig function, shifted horizontally. Or we can convert it to a double-angle equation.
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horizontal shift
We recall the sum of angles formula is ...
sin(x+y) = sin(x)cos(y) +cos(x)sin(y)
Using this, we can rewrite the left side of the equation using some angle offset y, and some scale factor k.
k·sin(x +y) = k·sin(y)·cos(x) +k·cos(y)·sin(x) = 2cos(x) +2sin(x)
finding the shift
Equating coefficients of cos(x) and sin(x), we can solve for k and y:
k·sin(y) = 2
k·cos(y) = 2
The ratio of these equations is ...
(k·sin(y))/(k·cos(y)) = 2/2
tan(y) = 1 ⇒ y = π/4
k = 2/sin(π/4) = 2√2
shifted equation
So, our original equation becomes ...
2√2·sin(x +π/4) = √6
Dividing by √2 and using the inverse sine function, we have ...
x +π/4 = arcsin((√3)/2) = π/2 ±π/6
x = π/4 ±π/6
x = π/12, 5π/12
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double-angle equation
If we square both sides of the original equation, we get ...
(2sin(x) +2cos(x))² = (√6)²
4sin²(x) +8sin(x)cos(x) +4cos²(x) = 6
2sin(x)cos(x) = (6 -4)/4 = 1/2 . . . . use sin² +cos² = 1, subtract 4, divide by 4
Using the trig identity 2sin(x)cos(x) = sin(2x), we can find ...
2x = arcsin(1/2) = π/2 ±π/3 +2kπ . . . . k = an integer;
x = π/4 ±π/6 +kπ . . . . for integer k
x = {π/12, 5π/12, 13π/12, 17π/12}
We know that squaring the equation can introduce extraneous solutions, so we need to try these out. For x in the third quadrant, the sine and cosine values are both negative, so the only useful solutions here are ...
x = π/12, 5π/12
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Additional comment
Based on the above, we now know that any trig expression of the form ...
a·sin(x) +b·cos(x)
can be rewritten to the form ...
(√(a² +b²))·sin(x +arctan(b/a)) . . . . . a scaled and shifted sine function
The arctangent will have to take the signs of 'a' and 'b' into account in order to get the angle quadrant right.
