solve trigonometric function?

Question

cosx+sinx=1

Answer 1

Method 1
By inspection
cos 90° + sin 90° = 0 + 1 = 1
Thus x = 90° is an answer.
Also:-
cos 0° + sin 0° = 1 + 0 = 1
Thus x = 0 ° is an answer.
Also:-
cos 360° + sin 360° = 1 + 0 = 1
Thus x = 360° is an answer.

Method 2
Let cos x + sin x = k cos (x - a) where k > 0
cos x + sin x = (k cos a) cos x + (k sin a) sin x
1 = k cos a
1 = k sin a (and cos a and sin a are + ve)
angle a is in 1st quadrant.
k sin a / k cos a = 1
tan a = 1
a = 45°

1² + 1² = k² (cos²a + sin²a)
2 = k²
k = √2
√2 cos (x - 45)° = 1
cos (x - 45)° = 1 / √2
x - 45 = 45 , 315°
x = 90° , 360° (and 0°) are answers.

Answer 2

One of the formal methods for solving equations of the type:
a cos(x) + b sin(x) = c
is to divide by m = sqrt(a^2 + b^2) getting:
(a/m)cos(x) + (b/m)sin(x) = c/m ....(1)
Because (a/m)^2 + (b/m)^2 = 1, there is an angle t such that cos(t) = a/m and sin(t) = b/m ......(2)
(1) can then be written:
cos(t)cos(x) + sin(t)sin(x) = c/m
cos(x - t) = c/m .........(3)
For this example, with a = b = c = 1, m = sqrt(2), making
t = pi/4, and (3) becomes:
cos(x - pi/4) = 1/sqrt(2)
Hence:
x - pi/4 = 2n pi +/- pi/4 where n is any integer
x = 2n pi +/- pi/2
x = n pi/2.

Answer 3

cosx=1-sinx

(1-sinx)^2+sinx^2=1 (because cosx^2+sinx^2=1)

1-2sinx+2sinx^2=1

2sinx(sinx-1)=0.

So sinx=0 -> cosx=1 true for 0, 360, 720 degrees
or sinx=1 -> cosx=0 true for 90, 450...

So x=n*2pi or x=pi/2+n*2pi=pi(1+4n)/2

Answer 4

first isolate cos cos(x - 40 5) = 3/4 your instructor suggested that because of the fact 3/4 isn't a 'established' element on the unit circle (for basic computation), i.e. from a 30-60-ninety triangle or 40 5-40 5-ninety triangle. rather, the customary factors would be from a multiple of 30 ranges, or a multiple of 40 5 ranges. and, if use cos(a - b) = cos(a)cos(b) + sin(a)sin(b) (or any others) it may render you in a similar subject. cos(x - 40 5) = cos(x)cos(40 5) + sin(x)sin45) = 3/4 sqrt(2) / 2 * (cos(x) + sin(x)) = 3/4 cos(x) + sin(x) = 3sqrt(2) / 8 squaring the two section: cos^2(x) + 2cos(x)sin(x) + sin^2(x) = 9/sixteen 2cos(x)sin(x) + a million = 9/sixteen sin(2x) + a million = 9/sixteen sin(2x) = - 7/sixteen u in basic terms wouldnt have the skill to ever get the customary factors. yet once you're rather in touch contained interior the set of rules to choose for non-established trig factors, its performed with tables of logarithms, i endure in techniques.

Answer 5

When x = 0, cos(x)=1 and sin(x)=0, so they add to 1.
When x = π/2, cos(x)=0 and sin(x)=1, so they add to 1.
since this is the max val of the func. provided therefore u have got the two answers.

The solutions are x = npi,where n is any integer.

Answer 6

When x = 0, cos(x)=1 and sin(x)=0, so they add to 1.
When x = π/2, cos(x)=0 and sin(x)=1, so they add to 1.

The solutions are
x = (4*n)*π/2
and
x = (4*n + 1)*π/2
where n = any integer

Or, you can think of it like this:
x = n*π/2
where n = 0, 1, 4, 5, 8, 9, 12, 13...