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There are operators missing.
The only question that I can see to make sense is:
Solve x for: cos(5x + 49°) = sin(3x + 57°)
It's been a while since I did this kind of problem, so there may be more solutions to the ones I give here:
Cosθ = sin(90 - θ)
→ cos(5x + 49°) = sin(3x + 57°)
→ sin(90° - (5x + 49°)) = sin(3x + 57°)
→ sin(41° - 5x) = sin(3x + 57°)
Thus:
41° - 5x = 3x + 57°
→ 8x = -16°
→ x = -2°
But as sin and cos are cyclic with a period of 360°, -2° = 360° - 2= 358°
→ x = 358° + 360°n where n = 0, 1, 2, ....
But sin θ = sin(180° - θ) which means that
180 - (41° - 5x) = 3x + 57°
→ 5x + 139° = 3x + 57°
→ 2x = -82°
→ x = -41°
→ x = 319° + 360°n where n = 0, 1, 2, 3,... is also a solution set.
Thus the solutions are:
x = 358° + 360°n
x = 319° + 360°n
where n = 0, 1, 2, 3, ...
What else can I help you with?
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It is equivalent to: 3x+6+5 = 3x+11
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Find the derivatives of y equals 2 sin 3x?
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Sin2x equals cos3x find x?
0. sin 2x = cos 3x 1. sin 2x = sin (pi/2 - 3x) [because cos u = sin (pi/2 - u)] 2. [...]
How would you evaluate the indefinite integral -2xcos3xdx?
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Integration cos3 x?
Take f(x) = cos(3x) ∫ f(x) dx = ∫ cos(3x) dx Take u=3x → du = 3dx = ∫ 1/3*cos(u) du = 1/3*∫ cos(u) du = 1/3*sin(u) + C, C ∈ ℝ = 1/3*sin(3x) + C
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integral sin(3 x) cos(5 x) dx = 1/16 (8 cos^2(x)-cos(8 x))+C
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convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx
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How do you solve Sin x plus sin 3x plus sin 5x plus sin 7x equals?
sin7x-sin6x+sin5x
