Solve the following trigonometric equations for their general solutions: a) sin 2x + cos x = 0 b) 6 cos2 x + sin x - 5 = 0
|Mathematics
Solve the following trigonometric equations for their general solutions: a) sin 2x + cos x = 0 b) 6 cos2 x + sin x - 5 = 0
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Answer
x=90∘+k⋅180∘orx=210∘+k⋅360∘orx=330∘+k⋅360∘,k∈Z
You're on a roll —
To determine the general solution for each equation, we will use trigonometric identities and the general solution formulas.
a) sin2x+cosx=0
Step 1: Use the double angle identity sin2x=2sinxcosx.
2sinxcosx+cosx=0
Step 2: Factor out cosx.
cosx(2sinx+1)=0
Step 3: Set each factor equal to zero and solve for x.
Case 1: cosx=0
The reference angle for cosx=0 is 90∘.
The general solution for cosx=0 is:
x=90∘+k⋅180∘,k∈Z
Case 2: 2sinx+1=02sinx=−1sinx=−21
The reference angle for sinx=21 is 30∘.
Since sinx is negative, x lies in the 3rd or 4th quadrant.
For the 3rd quadrant:
x=180∘+30∘+k⋅360∘x=210∘+k⋅360∘,k∈Z
For the 4th quadrant:
x=360∘−30∘+k⋅360∘x=330∘+k⋅360∘,k∈Z
The general solutions for (a) are:
x=90∘+k⋅180∘orx=210∘+k⋅360∘orx=330∘+k⋅360∘,k∈Z
b) 6cos2x+sinx−5=0
Step 1: Use the identity cos2x=1−sin2x to express the equation in terms of sinx.
6(1−sin2x)+sinx−5=06−6sin2x+sinx−5=0−6sin2x+sinx+1=0
Step 2: Multiply by -1 to make the leading coefficient positive.
6sin2x−sinx−1=0
Step 3: Let y=sinx. The equation becomes a quadratic equation: 6y2−y−1=0.
Factor the quadratic equation:
(3y+1)(2y−1)=0
Step 4: Substitute back sinx for y and solve for sinx.
Case 1: 3sinx+1=03sinx=−1sinx=−31
The reference angle is ref=arcsin(31)≈19.47∘.
Since sinx is negative, x lies in the 3rd or 4th quadrant.
For the 3rd quadrant:
x=180∘+19.47∘+k⋅360∘x=199.47∘+k⋅360∘,k∈Z
For the 4th quadrant:
x=360∘−19.47∘+k⋅360∘x=340.53∘+k⋅360∘,k∈Z
Case 2: 2sinx−1=02sinx=1sinx=21
The reference angle is ref=30∘.
Since sinx is positive, x lies in the 1st or 2nd quadrant.
For the 1st quadrant:
x=30∘+k⋅360∘,k∈Z
For the 2nd quadrant:
x=180∘−30∘+k⋅360∘x=150∘+k⋅360∘,k∈Z
The general solutions for (b) are:
x=30∘+k⋅360∘orx=150∘+k⋅360∘orx=199.47∘+k⋅360∘orx=340.53∘+k⋅360∘,k∈Z
c) cos2x−7cosx−3=0
Step 1: Use the double angle identity cos2x=2cos2x−1 to express the equation in terms of cosx.
(2cos2x−1)−7cosx−3=02cos2x−7cosx−4=0
Step 2: Let y=cosx. The equation becomes a quadratic equation: 2y2−7y−4=0.
Factor the quadratic equation:
(2y+1)(y−4)=0
Step 3: Substitute back cosx for y and solve for cosx.
Case 1: 2cosx+1=02cosx=−1cosx=−21
The reference angle is ref=arccos(21)=60∘.
Since cosx is negative, x lies in the 2nd or 3rd quadrant.
For the 2nd quadrant:
x=180∘−60∘+k⋅360∘x=120∘+k⋅360∘,k∈Z
For the 3rd quadrant:
x=180∘+60∘+k⋅360∘x=240∘+k⋅360∘,k∈Z
Case 2: cosx−4=0cosx=4
This case has no solution because the range of cosx is [−1,1].
The general solutions for (c) are:
x=120∘+k⋅360∘orx=240∘+k⋅360∘,k∈Z
d) sin2x=cos(x−30∘)
Step 1: Convert cos(x−30∘) to a sine function using the co-function identity cosθ=sin(90∘−θ).
cos(x−30∘)=sin(90∘−(x−30∘))cos(x−30∘)=sin(90∘−x+30∘)cos(x−30∘)=sin(120∘−x)
Step 2: Now the equation is sin2x=sin(120∘−x).
Apply the general solution for sinA=sinB:
A=B+k⋅360∘ or A=180∘−B+k⋅360∘.
Case 1: 2x=(120∘−x)+k⋅360∘2x+x=120∘+k⋅360∘3x=120∘+k⋅360∘x=3120∘+3k⋅360∘x=40∘+k⋅120∘,k∈Z
Case 2: 2x=180∘−(120∘−x)+k⋅360∘2x=180∘−120∘+x+k⋅360∘2x=60∘+x+k⋅360∘2x−x=60∘+k⋅360∘x=60∘+k⋅360∘,k∈Z
The general solutions for (d) are:
x=40∘+k⋅120∘orx=60∘+k⋅360∘,k∈Z
e) cos(x−30∘)=2sinx
Step 1: Use the compound angle identity cos(A−B)=cosAcosB+sinAsinB.
cosxcos30∘+sinxsin30∘=2sinx
Step 2: Substitute the exact values for cos30∘=23 and sin30∘=21.
cosx(23)+sinx(21)=2sinx
Step 3: Multiply the entire equation by 2 to eliminate fractions.
3cosx+sinx=4sinx
Step 4: Rearrange the terms to isolate sinx and cosx.
3cosx=4sinx−sinx3cosx=3sinx
Step 5: Divide both sides by cosx (assuming cosx=0) and by 3 to get tanx.
33=cosxsinxtanx=33
Step 6: Find the general solution for tanx=33.
The reference angle for tanx=33 is 30∘.
The general solution for tanx=tanα is x=α+k⋅180∘.
x=30∘+k⋅180∘,k∈Z
The general solution for (e) is:
x=30∘+k⋅180∘,k∈Z
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This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
You're on a roll — To determine the general solution for each equation, we will use trigonometric identities and the general solution formulas. a) 2x + x = 0 Step 1: Use the double angle identity 2x = 2 x x. 2 x x + x = 0 Step 2: Factor out x. x (2 x + 1) = 0 Step 3: Set each factor equal to zero and solve for x. Case 1: x = 0 The reference angle for x = 0 is 90^. The general solution for x = 0 is: x = 90^ + k · 180^, k Z Case 2: 2 x + 1 = 0 2 x = -1 x = -(1)/(2) The reference angle for x = (1)/(2) is 30^. Since x is negative, x lies in the 3rd or 4th quadrant. For the 3rd quadrant: x = 180^ + 30^ + k · 360^ x = 210^ + k · 360^, k Z For the 4th quadrant: x = 360^ - 30^ + k · 360^ x = 330^ + k · 360^, k Z The general solutions for (a) are: x = 90^ + k · 180^ or x = 210^ + k · 360^ or x = 330^ + k · 360^, k Z b) 6 ^2 x + x - 5 = 0 Step 1: Use the identity ^2 x = 1 - ^2 x to express the equation in terms of x. 6(1 - ^2 x) + x - 5 = 0 6 - 6 ^2 x + x - 5 = 0 -6 ^2 x + x + 1 = 0 Step 2: Multiply by -1 to make the leading coefficient positive. 6 ^2 x - x - 1 = 0 Step 3: Let y = x. The equation becomes a quadratic equation: 6y^2 - y - 1 = 0. Factor the quadratic equation: (3y + 1)(2y - 1) = 0 Step 4: Substitute back x for y and solve for x. Case 1: 3 x + 1 = 0 3 x = -1 x = -(1)/(3) The reference angle is ref = ((1)/(3)) ≈ 19.47^. Since x is negative, x lies in the 3rd or 4th quadrant. For the 3rd quadrant: x = 180^ + 19.47^ + k · 360^ x = 199.47^ + k · 360^, k Z For the 4th quadrant: x = 360^ - 19.47^ + k · 360^ x = 340.53^ + k · 360^, k Z Case 2: 2 x - 1 = 0 2 x = 1 x = (1)/(2) The reference angle is ref = 30^. Since x is positive, x lies in the 1st or 2nd quadrant. For the 1st quadrant: x = 30^ + k · 360^, k Z For the 2nd quadrant: x = 180^ - 30^ + k · 360^ x = 150^ + k · 360^, k Z The general solutions for (b) are: x = 30^ + k · 360^ or x = 150^ + k · 360^ or x = 199.47^ + k · 360^ or x = 340.53^ + k · 360^, k Z c) 2x - 7 x - 3 = 0 Step 1: Use the double angle identity 2x = 2 ^2 x - 1 to express the equation in terms of x. (2 ^2 x - 1) - 7 x - 3 = 0 2 ^2 x - 7 x - 4 = 0 Step 2: Let y = x. The equation becomes a quadratic equation: 2y^2 - 7y - 4 = 0. Factor the quadratic equation: (2y + 1)(y - 4) = 0 Step 3: Substitute back x for y and solve for x. Case 1: 2 x + 1 = 0 2 x = -1 x = -(1)/(2) The reference angle is ref = ((1)/(2)) = 60^. Since x is negative, x lies in the 2nd or 3rd quadrant. For the 2nd quadrant: x = 180^ - 60^ + k · 360^ x = 120^ + k · 360^, k Z For the 3rd quadrant: x = 180^ + 60^ + k · 360^ x = 240^ + k · 360^, k Z Case 2: x - 4 = 0 x = 4 This case has no solution because the range of x is [-1, 1]. The general solutions for (c) are: x = 120^ + k · 360^ or x = 240^ + k · 360^, k Z d) 2x = (x - 30^) Step 1: Convert (x - 30^) to a sine function using the co-function identity = (90^ - ). (x - 30^) = (90^ - (x - 30^)) (x - 30^) = (90^ - x + 30^) (x - 30^) = (120^ - x) Step 2: Now the equation is 2x = (120^ - x). Apply the general solution for A = B: A = B + k · 360^ or A = 180^ - B + k · 360^. Case 1: 2x = (120^ - x) + k · 360^ 2x + x = 120^ + k · 360^ 3x = 120^ + k · 360^ x = (120^)/(3) + (k · 360^)/(3) x = 40^ + k · 120^, k Z Case 2: 2x = 180^ - (120^ - x) + k · 360^ 2x = 180^ - 120^ + x + k · 360^ 2x = 60^ + x + k · 360^ 2x - x = 60^ + k · 360^ x = 60^ + k · 360^, k Z The general solutions for (d) are: x = 40^ + k · 120^ or x = 60^ + k · 360^, k Z e) (x - 30^) = 2 x Step 1: Use the compound angle identity (A - B) = A B + A B. x 30^ + x 30^ = 2 x Step 2: Substitute the exact values for 30^ = sqrt(3)2 and 30^ = (1)/(2). x (sqrt(3)2) + x ((1)/(2)) = 2 x Step 3: Multiply the entire equation by 2 to eliminate fractions. sqrt(3) x + x = 4 x Step 4: Rearrange the terms to isolate x and x. sqrt(3) x = 4 x - x sqrt(3) x = 3 x Step 5: Divide both sides by x (assuming x ≠ 0) and by 3 to get x. sqrt(3)3 = ( x)/( x) x = sqrt(3)3 Step 6: Find the general solution for x = sqrt(3)3. The reference angle for x = sqrt(3)3 is 30^. The general solution for x = is x = + k · 180^. x = 30^ + k · 180^, k Z The general solution for (e) is: x = 30^ + k · 180^, k Z What's next?
