find all solutions of 2sinx=1-cosx in the interval from 0 to 360
square both sides.
4sin^2= 1-2cos+cos^2
4(1-cos^2)=1-2cos+cos^2
now, gather terms, it is a quadratic, solve using the quadratic equation.
okay so i got -(3/5) and 1, i am supposed to find the solutions in the interval from 0degrees to 360degrees
the previous ans. is wrong as if we sq. the eqn. it generates false solutions a better method is to divide coefficients of sin and cos by root of sq. of coeff.ofsin+sq. of coeff of cos . that is main pt. remaining we will proceed by making it an argument in cos . and then make general sol. and find no. of sol. or diff solutions by putting value of n in general solution
the previous ans. is wrong as if we sq. the eqn. it generates false solutions a better method is to divide coefficients of sin and cos by root of( sq. of coeff.ofsin+sq. of coeff of cos) . that is main pt. remaining we will proceed by making it an argument in cos . and then make general sol. and find no. of sol. or diff solutions by putting value of n in general solution
To find all the solutions of the equation 2sinx = 1 - cosx in the interval from 0 to 360, we can use trigonometric identities and algebraic manipulations.
Step 1: Rewrite the equation using trigonometric identities:
2sinx = 1 - cosx
Step 2: Expand the right side using the identity cos(x) = 1 - 2sin^2(x):
2sinx = 1 - (1 - 2sin^2x)
2sinx = 2sin^2x
Step 3: Rearrange the equation:
2sin^2x - 2sinx + 1 = 0
Step 4: Solve the quadratic equation:
To solve this quadratic equation, we can use the quadratic formula, which is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = -2, and c = 1. Substituting these values into the quadratic formula:
x = (-(-2) ± √((-2)^2 - 4(2)(1))) / (2(2))
x = (2 ± √(4 - 8)) / 4
x = (2 ± √(-4)) / 4
x = (2 ± 2i) / 4
Step 5: Simplify the complex roots:
Divide both the numerator and denominator by 2 to simplify the complex roots:
x = 1/2 ± (1/2)i
So, the complex solutions in the interval from 0 to 360 are:
x = 1/2 + (1/2)i and x = 1/2 - (1/2)i
However, since the original equation involves sine and cosine functions, we need to find the real solutions within the given interval.
Step 6: Use the fact that sin(x) = cos(90 - x):
Substitute sinx = cos(90 - x) into the equation:
2cos(90 - x) = 1 - cosx
Step 7: Simplify the equation:
2cos(90 - x) + cosx = 1
Step 8: Expand cos(90 - x) using the identity cos(a + b) = cos(a)cos(b) - sin(a)sin(b):
2(cos90cosx + sin90sinx) + cosx = 1
Step 9: Simplify further:
2(0 + sinx) + cosx = 1
2sinx + cosx = 1
Step 10: Rearrange the equation:
cosx = 1 - 2sinx
Step 11: Square both sides of the equation to eliminate the square root:
cos^2x = (1 - 2sinx)^2
1 - sin^2x = 1 - 4sinx + 4sin^2x
4sin^2x - 5sinx + 1 = 0
Step 12: Factor the quadratic equation:
(4sinx - 1)(sinx - 1) = 0
Step 13: Set each factor equal to zero and solve for x:
4sinx - 1 = 0 --> sinx = 1/4
sinx - 1 = 0 --> sinx = 1
Step 14: Find the corresponding angles:
Using the unit circle or a calculator, we can find the angles that correspond to sin(1/4) and sin(1).
sin(1/4) ≈ 0.2588, which gives two possible solutions:
x = arcsin(0.2588) ≈ 0.268 radians ≈ 15.34 degrees
or
x = 180 - arcsin(0.2588) ≈ 164.66 degrees
sin(1) ≈ 0.8415, which gives one solution:
x = arcsin(0.8415) ≈ 1.14 radians ≈ 65.57 degrees
So, the real solutions in the interval from 0 to 360 are:
x ≈ 0.268 radians ≈ 15.34 degrees,
x ≈ 164.66 degrees,
and x ≈ 65.57 degrees.