1)Solve the equation: sin(2x+10)°=√3/2 for 0°

Question

1)Solve the equation: sin(2x+10)°=√3/2 for 0°<or=x<or=360°
2)Solve for x in the equation:sin[5/2x]°=-1/2 for 0°<or=x<or<or=360°
3)Solve for x in the equation; tan x=2sin x for -180°<or=x<or=180°
4)Find the value of x in the equation; cos(3x+180)°=√3/2 for 0°<or=x<or=180°
5)Solve for titre in the equation; -1/4sin(2titre+30)°=0.1607 for 0°<or= x<or=360°
6)Given that cos a=5/13 and that 270°<or=a<or=360°,evaluate tan a+sin a
7)If tan titre=12/5 and titre is reflex,find the value of 5sin titre+cos titre
8)Find titre given that 2cos3titre-1=0 for 0°<or=x<or=360°
9)Solve the equation; 2cos2(x+30)°=-1 for 0°<or=x<or=360°
10)Solve for titre in the equation; cos(2titre+30)=-0.7660 for -180°<or=x<or=180°
11)Solve the equation 2 tan 2xcos2x+√3=0 for  -180°<or=x<or=180°
12)Given that 3tan^2(2x+2/3π°=1,Find x when 0°<or=x<or=2π°.Give your answer in π°
13)A figure has point D which divides AC in the ratio 1:1.AB=6.4cm, angle ABD=39° and angle BAD=78°.Calculate correct to 1 d.p;
a)The length of BD
b)The length of BC
14)A triangle ABC is inscribed in a circle.AB=6cm.BC=9cm and angle ABC=80°.Calculate to one decimal place;
a)The length of AC
b)The radius of the circle
c)The area of the shaded segments

GPT 3.5 · Answer

1) To solve the equation sin(2x+10)° = √3/2 for 0° ≤ x ≤ 360°, we need to find the values of x that make the equation true.

First, consider when sin(2x+10)° = √3/2. This occurs when the angle (2x+10)° is equal to π/3 or 60°.

Now we can set up an equation: 2x + 10 = 60.

Subtracting 10 from both sides gives us: 2x = 50.

Dividing both sides by 2 gives us: x = 25.

So the solution to the equation is x = 25.

2) To solve the equation sin(5/2x)° = -1/2 for 0° ≤ x ≤ 360°, we need to find the values of x that make the equation true.

First, consider when sin(5/2x)° = -1/2. This occurs when the angle (5/2x)° is equal to -π/6 or -30°.

Now we can set up an equation: 5/2x = -30.

Multiplying both sides by 2/5 gives us: x = -12.

So the solution to the equation is x = -12.

3) To solve the equation tan(x)=2sin(x) for -180° ≤ x ≤ 180°, we need to find the values of x that make the equation true.

First, we can rewrite tan(x) as sin(x)/cos(x). Now the equation becomes sin(x)/cos(x) = 2sin(x).

Multiplying both sides by cos(x) gives us: sin(x) = 2sin(x)cos(x).

Dividing both sides by sin(x) gives us: 1 = 2cos(x).

Dividing both sides by 2 gives us: 1/2 = cos(x).

So the solution to the equation is x = 60°.

4) To solve the equation cos(3x+180)° = √3/2 for 0° ≤ x ≤ 180°, we need to find the values of x that make the equation true.

First, consider when cos(3x+180)° = √3/2. This occurs when the angle (3x+180)° is equal to π/6 or 30°.

Now we can set up an equation: 3x + 180 = 30.

Subtracting 180 from both sides gives us: 3x = -150.

Dividing both sides by 3 gives us: x = -50.

So the solution to the equation is x = -50.

5) To solve the equation -1/4sin(2t+30)° = 0.1607 for 0° ≤ t ≤ 360°, we need to find the values of t that make the equation true.

First, consider when -1/4sin(2t+30)° = 0.1607. This occurs when the angle (2t+30)° is equal to sin^(-1)(-0.1607) or -9.209°.

Now we can set up an equation: 2t + 30 = -9.209.

Subtracting 30 from both sides gives us: 2t = -39.209.

Dividing both sides by 2 gives us: t = -19.6045.

So the solution to the equation is t = -19.6045.