Sinx+root3cos=1 if 0<=x<360
Root3cosx nt root3cos
If you meant
sinx + √3 cosx = 1
then try this
√3 cosx = 1-sinx
3cos^2x = 1-2sinx+sin^2x
3(1-sin^2x) = 1-2sinx+sin^2x
4sin^2x - 2sinx - 2 = 0
2(2sinx+1)(sinx-1) = 0
sinx = -1/2 or 1
x = 90 or 210 or 330
But 210 does not work in the original equation, so throw it out.
sinx+1.732cosx=1
2(sinx(1/2)+(1.732/2)cosx)=1
sin(x+pi/3)=1/2
x+pi/3=5pi/6
x=3pi/2
To solve the equation sin(x) + √3cos(x) = 1, we can use trigonometric identities and properties to simplify the equation and find the values of x that satisfy the equation. Here's how we can approach it:
1. Convert the equation to a single trigonometric function:
Use the identity cos(x) = sin(π/2 - x) to rewrite the equation.
sin(x) + √3cos(x) = sin(x) + √3sin(π/2 - x)
2. Apply the sum-to-product identities:
sin(a) + sin(b) = 2sin((a + b)/2)cos((a - b)/2)
sin(x) + √3sin(π/2 - x) = 2sin((x + π/2 - x)/2)cos((x + π/2 + x)/2)
= 2sin(π/4)cos((π/2 + x)/2)
3. Simplify further:
sin(π/4) = √2/2
2sin(π/4)cos((π/2 + x)/2) = √2cos((π/2 + x)/2)
4. Apply the double-angle identity:
cos(2θ) = 2cos^2(θ) - 1
cos((π/2 + x)/2) = √2cos((π + 2x)/4) = √2[2cos^2((π + 2x)/4) - 1]
Now we have simplified the original equation sin(x) + √3cos(x) = 1 to:
√2[2cos^2((π + 2x)/4) - 1] = 1
5. Solve for cos((π + 2x)/4):
√2[2cos^2((π + 2x)/4) - 1] = 1
2cos^2((π + 2x)/4) - 1 = 1/√2
2cos^2((π + 2x)/4) = 1/√2 + 1
cos^2((π + 2x)/4) = (1/√2 + 1)/2
cos((π + 2x)/4) = ±√[((1/√2 + 1)/2)]
6. Solve for (π + 2x)/4:
(π + 2x)/4 = ±arccos(√[((1/√2 + 1)/2)])
π + 2x = ±4arccos(√[((1/√2 + 1)/2)])
x = (±4arccos(√[((1/√2 + 1)/2)]) - π)/2
Now you can solve for x by substituting values for (±4arccos(√[((1/√2 + 1)/2)]) - π)/2 depending on the range of x values you're considering (0 <= x < 360).